ENGINEERING DYNAMICS
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ENGINEERING DYNAMICS
A Comprehensive Introduction
N. Jeremy Kasdin and Derek A. Paley
PRINCETON UNIVERSITY PRESS
PRINCETON AND OXFORD
Copyright © 2011 by Princeton University Press
Published by Princeton University Press, 41 William Street, Princeton,
New Jersey 08540
In the United Kingdom: Princeton University Press, 6 Oxford Street, Woodstock,
Oxfordshire OX20 1TW
press.princeton.edu
All Rights Reserved
Library of Congress Cataloging-in-Publication Data
Kasdin, N. Jeremy.
Engineering dynamics : a comprehensive introduction / N. Jeremy Kasdin and
Derek A. Paley.
p. cm.
Includes bibliographical references and index.
ISBN 978-0-691-13537-3 (hardback : alk. paper) 1. Dynamics. 2. Mechanics.
I. Paley, Derek A., 1974– II. Title.
TA352.K375 2010
620.1 04—dc22
2010023802
British Library Cataloging-in-Publication Data is available
This book has been composed in Times Roman and Avenir using ZzTEX
by Princeton Editorial Associates, Inc., Scottsdale, Arizona.
Printed on acid-free paper.
Printed in the United States of America
10 9 8 7 6 5 4 3 2 1
To our wives Kef and Robyn and our children Alex, Izzy, Ethan,
and Adyn—your dynamics make it all worthwhile.
Philosophy is written in this grand book—I mean the universe—which stands
continually open to our gaze, but the book cannot be understood unless one first
learns to comprehend the language and read the letters in which it is composed. It is
written in the language of mathematics.
Galileo Galilei, The Controversy on the Comets of 1618
CONTENTS
Preface
xi
Chapter 1. Introduction
1.1
What Is Dynamics?
1.2
Organization of the Book
1.3
Key Ideas
1.4
Notes and Further Reading
1.5
Problems
1
1
6
8
9
10
Chapter 2. Newtonian Mechanics
2.1
Newton’s Laws
2.2
A Deeper Look at Newton’s Second Law
2.3
Building Models and the Free-Body Diagram
2.4
Constraints and Degrees of Freedom
2.5
A Discussion of Units
2.6
Tutorials
2.7
Key Ideas
2.8
Notes and Further Reading
2.9
Problems
11
11
15
19
21
24
25
37
38
38
PART ONE. PARTICLE DYNAMICS IN THE PLANE
Chapter 3. Planar Kinematics and Kinetics of a Particle
3.1
The Simple Pendulum
3.2
More on Vectors and Reference Frames
3.3
Velocity and Acceleration in the Inertial Frame
3.4
Inertial Velocity and Acceleration in a Rotating Frame
3.5
The Polar Frame and Fictional Forces
3.6
An Introduction to Relative Motion
3.7
How to Solve a Dynamics Problem
3.8
Derivations—Properties of the Vector Derivative
3.9
Tutorials
3.10 Key Ideas
3.11 Notes and Further Reading
3.12 Problems
45
45
47
56
66
79
83
87
88
93
100
101
102
Chapter 4. Linear and Angular Momentum of a Particle
4.1
Linear Momentum and Linear Impulse
4.2
Angular Momentum and Angular Impulse
4.3
Tutorials
4.4
Key Ideas
4.5
Notes and Further Reading
4.6
Problems
113
113
117
131
141
142
143
viii
CONTENTS
Chapter 5. Energy of a Particle
5.1
Work and Power
5.2
Total Work and Kinetic Energy
5.3
Work Due to an Impulse
5.4
Conservative Forces and Potential Energy
5.5
Total Energy
5.6
Derivations—Conservative Forces and Potential Energy
5.7
Tutorials
5.8
Key Ideas
5.9
Notes and Further Reading
5.10 Problems
148
148
153
158
159
169
172
173
179
180
181
PART TWO. PLANAR MOTION OF A MULTIPARTICLE SYSTEM
Chapter 6. Linear Momentum of a Multiparticle System
6.1
Linear Momentum of a System of Particles
6.2
Impacts and Collisions
6.3
Mass Flow
6.4
Tutorials
6.5
Key Ideas
6.6
Notes and Further Reading
6.7
Problems
189
189
205
220
228
235
237
237
Chapter 7. Angular Momentum and Energy of a Multiparticle System
7.1
Angular Momentum of a System of Particles
7.2
Angular Momentum Separation
7.3
Total Angular Momentum Relative to an Arbitrary Point
7.4
Work and Energy of a Multiparticle System
7.5
Tutorials
7.6
Key Ideas
7.7
Notes and Further Reading
7.8
Problems
245
245
252
259
263
274
285
287
288
PART THREE. RELATIVE MOTION AND RIGID-BODY
DYNAMICS IN TWO DIMENSIONS
Chapter 8. Relative Motion in a Rotating Frame
8.1
Rotational Motion of a Planar Rigid Body
8.2
Relative Motion in a Rotating Frame
8.3
Planar Kinetics in a Rotating Frame
8.4
Tutorials
8.5
Key Ideas
8.6
Notes and Further Reading
8.7
Problems
295
295
302
311
318
328
329
330
Chapter 9. Dynamics of a Planar Rigid Body
9.1
A Rigid Body Is a Multiparticle System
9.2
Translation of the Center of Mass—Euler’s First Law
9.3
Rotation about the Center of Mass— Euler’s Second Law
9.4
Rotation about an Arbitrary Body Point
337
337
340
343
360
CONTENTS
9.5
9.6
9.7
9.8
9.9
9.10
ix
Work and Energy of a Rigid Body
A Collection of Rigid Bodies and Particles
Tutorials
Key Ideas
Notes and Further Reading
Problems
368
376
385
394
397
398
PART FOUR. DYNAMICS IN THREE DIMENSIONS
Chapter 10. Particle Kinematics and Kinetics
in Three Dimensions
10.1 Two New Coordinate Systems
10.2 The Cylindrical and Spherical Reference Frames
10.3 Linear Momentum, Angular Momentum, and Energy
10.4 Relative Motion in Three Dimensions
10.5 Derivations—Euler’s Theorem and the Angular Velocity
10.6 Tutorials
10.7 Key Ideas
10.8 Notes and Further Reading
10.9 Problems
Chapter 11. Multiparticle and Rigid-Body Dynamics
in Three Dimensions
11.1 Euler’s Laws in Three Dimensions
11.2 Three-Dimensional Rotational Equations
of Motion of a Rigid Body
11.3 The Moment Transport Theorem and the Parallel
Axis Theorem in Three Dimensions
11.4 Dynamics of Multibody Systems in Three Dimensions
11.5 Rotating the Moment of Inertia Tensor
11.6 Angular Impulse in Three Dimensions
11.7 Work and Energy of a Rigid Body in Three Dimensions
11.8 Tutorials
11.9 Key Ideas
11.10 Notes and Further Reading
11.11 Problems
409
409
413
422
426
445
450
458
459
460
465
465
472
495
502
504
509
510
515
523
526
527
PART FIVE. ADVANCED TOPICS
Chapter 12. Some Important Examples
12.1 An Introduction to Vibrations and Linear Systems
12.2 Linearization and the Linearized Dynamics of an Airplane
12.3 Impacts of Finite-Sized Particles
12.4 Key Ideas
12.5 Notes and Further Reading
537
537
551
568
578
579
Chapter 13. An Introduction to Analytical Mechanics
13.1 Generalized Coordinates
13.2 Degrees of Freedom and Constraints
13.3 Lagrange’s Method
580
580
583
589
x
CONTENTS
13.4
13.5
13.6
Kane’s Method
Key Ideas
Notes and Further Reading
605
618
619
APPENDICES
Appendix A. A Brief Review of Calculus
A.1 Continuous Functions
A.2 Differentiation
A.3 Integration
A.4 Higher Derivatives and the Taylor Series
A.5 Multivariable Functions and the Gradient
A.6 The Directional Derivative
A.7 Differential Volumes and Multiple Integration
623
623
624
626
627
629
632
633
Appendix B. Vector Algebra and Useful Identities
B.1 The Vector
B.2 Vector Magnitude
B.3 Vector Components
B.4 Vector Multiplication
635
635
637
637
638
Appendix C. Differential Equations
C.1 What Is a Differential Equation?
C.2 Some Common ODEs and Their Solutions
C.3 First-Order Form
C.4 Numerical Integration of an Initial Value Problem
C.5 Using matlab to Solve ODEs
645
645
647
650
651
657
Appendix D. Moments of Inertia of Selected Bodies
660
Bibliography
Index
663
667
PREFACE
Dynamics is difficult. There is no getting around that. This is particularly true for
undergraduates just starting their engineering and science education, when they are
beginning to wrestle with the physics and mathematics needed to gain facility with
dynamics. We find that simply acknowledging this fact goes a long way toward
increasing confidence. Nevertheless, the pedagogical solution is not to simplify the
material to make it more manageable. Rather, we feel quite strongly that students are
best served by employing careful rigor and emphasizing deep understanding of the
concepts as well as by using precise mathematics. In this way, they are provided with
tools and concepts that will serve them throughout their educational and professional
careers. The proper response to the admitted difficulty of the subject is to slow
down the presentation, perhaps stretching it over multiple quarters or semesters, and
gradually building complexity rather than simplifying in a way that lacks rigor and
care. To that end, we have included extensive appendices covering the mathematical
skills needed to understand all material in the book.
Most students who will use this book have had an introduction to mechanics in
their freshman physics courses. It is our goal to reintroduce them to the material with
the added sophistication of vector calculus and differential equations. Our approach
to ensuring both understanding and confidence is to emphasize careful notation and
rigor. Although some students complain about the pedantry and others want to jump to
the end, it is our experience that the way to ensure competence is to enforce a rigorous
and careful problem-solving process. Unfortunately, too close an adherence to this
principle can lead to a course—and textbook—that is dry, uninviting, and presented
in a way that is inconsistent with how students learn. The challenge we undertook
in writing this book was to maintain rigor (and rigorous notation) while making the
material sufficiently approachable and informal that students will spend time reading
it and wrestling with it.
Certainly there are many good books available that treat the subject of dynamics
with complete rigor. We confess that we like a good number of them and are attracted
to the top-down approach of developing the material from first principles, starting
with geometry, moving on to fully three-dimensional vector kinematics, and then
continuing through particle and rigid-body dynamics. In fact, we use this approach in
our graduate classes, where we also include Lagrangian and Hamiltonian methods.
However, we have found that undergraduates (especially sophomores and juniors)
have difficulty learning the material this way. Rather, a bottom-up approach that
develops skills and techniques on simpler problems—without sacrificing rigor—and
gradually increases sophistication—without losing sight of the basic physics—seems
to best capitalize on the way these students learn. In that sense our approach can
be likened to learning to play a musical instrument. We begin with the essential
fundamentals and, through repeated problem solving (practice), develop “muscle
memory” as new and more difficult pieces are tackled. Yet the notations we use from
the beginning—the notes, chords, and time signatures—remain the same and return
again and again.
xii
PREFACE
We thus take a unique approach in this book. We introduce Newton’s laws and start
solving important problems even before beginning a discussion of vector kinematics.
We seek to maintain student interest and present key notations and skills in the
context of real problems. An overemphasis on the mathematics, without maintaining
a connection to the physical objectives, can cause confusion and diminish enthusiasm
among students. For this reason, in some chapters we defer more detailed or complex
derivations to the end of the chapter, so as not to interrupt the physical picture.
Kinematics is developed slowly, always in the context of dynamics problems. Yet we
insist on a very careful notation, inspired by Thomas Kane’s wonderful books. We
always specify reference frames and are careful to maintain the distinction between
vectors, components, and scalars. The emphasis on using and understanding reference
frames (and specifying the inertial frame when solving problems) is something we
are particularly wedded to and find lacking in many introductory dynamics texts. In
our experience, the best thing students can do to avoid errors and enhance learning
is be compulsive about notation from the start.
We also emphasize finding equations of motion. Before computers became commonplace, dynamics education (as reflected in older textbooks) tended to emphasize
finding accelerations and treating dynamics problems as slightly more complicated
statics problems. Dynamics, however, is about finding equations of motion and determining trajectories. We thus introduce students early on to the idea of using ordinary
differential equations to describe the motion of systems and to the use of a computer
to integrate these equations. Where possible, analytical solutions to the equations of
motion are presented.
We have made every effort to include examples spanning a range of difficulty and
covering the most important concepts and techniques. We have tried to connect the
examples to real physical systems. Certain examples regularly repeat throughout the
book, so that students can see how new concepts are used on familiar problems and
how new insights can be gained from increasingly sophisticated analysis.
Our approach of distinguishing examples from tutorials allows us to employ simple
problems to highlight specific ideas just after they are introduced (examples) while
reserving problems that synthesize many concepts for the end of the chapter (tutorials). Some tutorials can be quite difficult, and instructors may want to judiciously
select among them; however, we felt presenting a wide range of difficulty and depth
resulted in a text that may prove useful for years after the course is taken.
We have also chosen to adopt an informal conversational style. Although purists
may be put off by this tone in a technical work, our feedback from students—after
trying a number of different textbooks—is that they appreciate the approachability of
conversational writing and find the material more accessible. We directly address the
reader and attempt to guide him or her through the difficult task of learning dynamics.
Acknowledgments
We owe a debt of gratitude to many people who aided in myriad ways, both direct
and behind the scenes, to make this book a reality. First and foremost, we thank our
students, who eagerly engaged in learning the material and refining the text. Without
them, their curiosity, and their desire to learn, this text would not exist. We also
PREFACE
xiii
thank our colleagues at Princeton and Maryland, who proved willing to engage in
endless discussions about dynamics. In particular, we are grateful to Phil Holmes,
Sean Humbert, Naomi Leonard, Michael Littman, Clancy Rowley, Ben Shapiro, Rob
Stengel, and Bob Vanderbei for regularly teaching us new things.
Prof. Kasdin is also indebted to the many professors, colleagues, and friends at
Stanford whose knowledge of dynamics, wisdom, and incredible insight are captured
throughout the book, particularly Art Bryson, Bob Cannon, Dan DeBra, Tom Kane,
Brad Parkinson, and Steve Rock.
We thank the many teaching assistants who worked above and beyond to make
sure the book was correct, that examples worked, and that problems had solutions.
They include Kevin Anderson, Tyler Groff, Ben Jorns, Jason Kay, Adele Lim, Ben
Nabet, Laurent Pueyo, Harinder Singh, Andy Stewart, and Nitin Sydney. We can’t
thank Dmitry Savransky enough; beyond being an amazing teaching assistant, he has
put in countless hours creating figures, examples, and problems and supporting almost
every aspect of the book’s production. Without him, the book would not be what it is.
We thank the manuscript reviewers, whose thorough reading and insightful suggestions improved the manuscript enormously. We thank John Lienhard for generously
providing his images of the Watt flyball governor. We thank Peter Strupp, Cyd Westmoreland, and the staff at Princeton Editorial Associates, whose tireless editing made
this a book worth reading, and Mark Bellis at Princeton University Press, who shepherded the book through its final stages. We are especially grateful to our editor, Ingrid
Gnerlich, who showed more patience than anyone should expect in guiding us through
the process of writing, editing, and production. She was always uplifting and always
encouraging.
Most importantly, we thank our families: Kef, Alex, Izzy, Robyn, Ethan, and Adyn.
Their love and support through the many days and nights we worked on the book made
this text possible.
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ENGINEERING DYNAMICS
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CHAPTER ONE
Introduction
1.1 What Is Dynamics?
Dynamics is the science that describes the motion of bodies. Also called mechanics
(we use the terms interchangeably throughout the book), its development was the first
great success of modern physics. Much notation has changed, and physics has grown
more sophisticated, but we still use the same fundamental ideas that Isaac Newton
developed more than 300 years ago (using the formulation provided by Leonhard
Euler and Joseph Louis Lagrange). The basic mathematical formulation and physical
principles have stood the test of time and are indispensable tools of the practicing
engineer.
Let’s be more precise in our definition. Dynamics is the discipline that determines
the position and velocity of an object under the action of forces. Specifically, it is about
finding a set of differential equations that can be solved (either exactly or numerically
on a computer) to determine the trajectory of a body.
In only the second paragraph of the book we have already introduced a great
number of terms that require careful, mathematical definitions to proceed with the
physics and eventually solve problems (and, perhaps, understand our admittedly very
qualitative definition): position, velocity, orientation, force, object, body, differential
equation, and trajectory. Although you may have an intuitive idea of what some of
these terms represent, all have rigorous meanings in the context of dynamics. This
rigor—and careful notation—is an essential part of the way we approach the subject
of dynamics in this book. If you find some of the notation to be rather burdensome
and superfluous early on, trust us! By the time you reach Part Two, you will find it
indispensable.
We begin in this chapter and the next by providing qualitative definitions of the
important concepts that introduce you to our notation, using only relatively simple
ideas from geometry and calculus. In Chapter 3, we are much more careful and present
the precise mathematical definitions as well as the full vector formulation of dynamics.
2
CHAPTER ONE
Q
P
rQ/P
rQ/O
rP/O
P
rP/O
O
O
(a)
(b)
Figure 1.1 (a) Vector rP/O from reference point O to point P represents the position of the
point P relative to O. (b) The addition of two vectors, rP/O and rQ/P , to get the resultant vector
rQ/O .
1.1.1 Vectors
We live in a three-dimensional Euclidean1 universe; we can completely locate the
position of a point P relative to a reference point O in space by its relative distance in
three perpendicular directions. (In Part One we talk about points rather than extended
bodies and, consequently, don’t have to keep track of the orientation of a body, as is
necessary when discussing rigid bodies in Parts Three and Four.) We often call the
reference point O the origin. An abstract quantity, the vector, is defined to represent
the position of P relative to O, both in distance and direction.
Qualitative Definition 1.1 A vector is a geometric entity that has both magnitude
and direction in space.2
A position vector is denoted by a boldface, roman-type letter with subscripts that
indicate its head and tail. For example, the position rP/O of point P relative to the
origin O is a vector (Figure 1.1). An important geometric property of vectors is that
they can be added to get a new vector, called the resultant vector. Figure 1.1b illustrates
how two vectors are added to obtain a new vector of different magnitude and direction
by placing the summed vectors “head to tail.”
When the position of point P changes with time, the position at time t is denoted by
rP/O (t). In this case, the velocity of point P with respect to O is also a vector. However,
to define the velocity correctly, we need to introduce the concept of a reference frame.
1.1.2 Reference Frames, Coordinates, and Velocity
We have all heard about reference frames since high school, and you may already have
an idea of what one is. For example, on a moving train, objects that are stationary on
the train—and thus with respect to a reference frame fixed to the train—move with
respect to a reference frame fixed to the ground (as in Figure 1.2). To successfully use
dynamics, such an intuitive understanding is essential. Later chapters discuss how
reference frames fit into the physics and how to use them mathematically; for that
1 Euclid
of Alexandria (ca. 325–265 BCE) was a Greek mathematician considered to be the father of
geometry. In his book The Elements, he laid out the basic foundations of geometry and the axiomatic
method.
2 In this book, a qualitative definition is typically followed by an operational or mathematical definition
of the same term, although the latter definition may come in a later chapter.
INTRODUCTION
3
B
I
Figure 1.2 Qualitative definition of a reference frame.
reason, we revisit the topic again in Chapter 3. For now, we summarize our intuition
in the following qualitative definition of a reference frame.
Qualitative Definition 1.2 A reference frame is a point of view from which
observations and measurements are made regarding the motion of a system.
It is impossible to overemphasize the importance of this concept. Solving a problem in dynamics always starts with defining the necessary reference frames.
From basic geometry, you may be used to seeing a reference frame written as
three perpendicular axes meeting at an origin O, as illustrated in Figure 1.3. This
representation is standard, as it highlights the three orthogonal Euclidean directions.
However, this recollection should not be confused with a coordinate system. The
reference frame and the coordinate system are not the same concept, but rather
complement one another. It is necessary to introduce the reference frame to define
a coordinate system, which we do next.
I
O
Figure 1.3 Reference frame I is represented by three mutually perpendicular axes meeting at
origin O.
4
CHAPTER ONE
I
P
I
d (r ) = Iv
—
P/O
P/O
dt
rP/O(t)
rP/O(t + Δt)
O
Figure 1.4 Velocity IvP/O is the instantaneous rate of change of position rP/O with respect to
frame I . That is, IvP/O = (rP/O (t + t) − rP/O (t))/t, in the limit t → 0.
Definition 1.3 A coordinate system is the set of scalars that locate the position of
a point relative to another point in a reference frame.
In our three-dimensional Euclidean universe, it takes three scalars to specify
the position of a point P in a reference frame. The most natural set of scalars
(the three numbers usually labeled x, y, and z) are Cartesian coordinates.3 These
coordinates represent the location of P in each of the three orthogonal directions of
the reference frame. (Recall the discussion of vectors in the previous section stating
that the position of P relative to O is specified in three perpendicular directions.)
Cartesian coordinates, however, are only one possible set of the many different scalar
coordinates, a number of which are discussed later in the book. Nevertheless, we
begin the study of dynamics with Cartesian coordinates because they have a one-toone correspondence with the directions of a reference frame. It is for this reason that
the Cartesian-coordinate directions are often thought to define the reference frame
(but don’t let this lure you into forgetting the distinction between a coordinate system
and a reference frame). We return to the concepts of reference frames and coordinate
systems and discuss the relationship between a coordinate system and a vector in
Chapter 3.
Throughout the book, reference frames are always labeled. Later we will be solving
problems that employ many different frames, and these labels will become very
important. Thus we often write the three Cartesian coordinates as (x, y, z)I , explicitly
noting the reference frame—here labeled I —in which the coordinates are specified.
(The reason for the letter I will become apparent later.)
Likewise, the change in time of a point’s position (i.e., the velocity) only has
meaning when referred to some reference frame (recall the train example). For that
reason, we always explicitly point out the appropriate reference frame when writing
the velocity. A superscript calligraphic letter is used to indicate the frame. Figure 1.4
Id
shows a schematic picture of the velocity IvP/O =
dt (rP/O ) as the instantaneous rate
of change in time of the position rP/O with respect to the frame I .4
We can also express the velocity of point P with respect to O as the rate of change
d
dx
dy
dz
(ẋ, ẏ, ż)I =
dt (x, y, z)I , where ẋ = dt , ẏ = dt , and ż = dt . (Appendix A reviews
3 Named after René Descartes (1596–1650), the celebrated French philosopher, who founded analytic
geometry and invented the notation.
4 In this book, the symbol =
denotes a definition as opposed to an equality.
INTRODUCTION
5
some basic rules of calculus if you are rusty.) Because the variables x, y, and z are
scalars, their time derivatives do not need a frame identification. We maintain the
notation (ẋ, ẏ, ż)I , however, to remind you that these three scalars are the rates of
change of the three position coordinates in frame I . The rate of change of a scalar
Cartesian coordinate is called the speed to distinguish it from the velocity. We return
to this topic and discuss it in depth and more formally in Chapter 3.
1.1.3 Equations of Motion
We now return to the definition of dynamics. Trajectory signifies the complete specification of the three positions and three speeds of a point in a reference frame as a function of time. It takes six quantities in our three-dimensional universe to completely
specify the motion of a point. This is not necessarily obvious. Why six quantities and
not three? Isn’t the position enough (since we can always find the velocity by differentiating)? The answer is no, because dynamics is about more than just specifying
the position and velocity. It is about finding equations, based on Newton’s laws, that
allow us to predict the complete trajectory of an object given only its state at a single
moment in time. By state we mean the three positions and three speeds of the point.
These six quantities, defined at a single moment in time, are called the initial conditions. The tools of dynamics allow us to find a set of differential equations that can
be solved—using these initial conditions—for the position and velocity at any later
time. These differential equations are called equations of motion.5
Definition 1.4 The equations of motion of a point are three second-order differential equations6 whose solution is the position and velocity of the point as a
function of time.
To see this a bit more clearly, imagine that we know the three position variables
x(t), y(t), and z(t) of a point in frame I at some time t and wish to know the
position some short time later, t + t. Without the velocity at t we are lost; the
point could move anywhere. However, with the three speeds ẋ(t), ẏ(t), and ż(t),
we know everything; the new position of the point in I is (x(t) + ẋ(t)t, y(t) +
ẏ(t)t, z(t) + ż(t)t)I . The equations of motion allow us to find the speeds at time
t + t. The six positions and speeds are sufficient to find the complete trajectory.
As an example, one of the simplest equations of motion is that for a mass on a
spring. The position of the mass is given by the Cartesian coordinate x, and the force
due to the spring is given by −kx (see Figure 2.7c). The position thus satisfies the
following second-order differential equation, obtained by equating the force with
the mass times acceleration and solving for the acceleration:
ẍ = −
k
x.
m
This differential equation is an equation of motion. Its solution gives x(t) and ẋ(t),
the trajectory of the mass point. Don’t worry if you didn’t follow how the equation
was obtained; that is covered in Chapter 2.
5 Appendix
6 Or,
C supplies a brief review of differential equations.
equivalently, six first-order differential equations.
6
CHAPTER ONE
Many equations of motion cannot be solved exactly; a computer is required to find
numerical trajectories. You will have an opportunity to do this many times in this
book. However, often we skip solving for the trajectory and find special solutions
or conditions on the states by setting the time equal to a specific value, finding
certain conditions on the forces, or setting the acceleration to a constant or zero
(sometimes called a steady state). One particularly useful such solution is known as
an equilibrium point. The mathematical details of equilibrium solutions are presented
in Chapter 12, but it is useful to have a qualitative understanding now, as we will be
finding equilibrium solutions of many systems here and there throughout the book.
Qualitative Definition 1.5 An equilibrium point of a dynamic system is a specific solution of the equations of motion in which the rates of change of the
states are all zero.
In other words, an equilibrium point is a configuration in which the system is
at rest. For the mass-spring system, for example, there is one equilibrium point,
which corresponds to the mass situated at precisely the rest length of the spring.
Mathematically, if x(t) is an equilibrium point, then ẋ(t) = 0 and ẍ(t) = 0. Thus
x(t) = x(0), where x(0) is the initial condition at time t = 0. So an equilibrium point
is a solution whose value over time remains equal to its initial value.
In summary, dynamics is about finding three second-order differential equations
that can be solved for the complete trajectory of an object. The equations can be
solved—using the six initial conditions—either analytically (by hand) or numerically
(by a computer). It is true that other scalar quantities can be used to specify the position
rather than Cartesian coordinates; we will begin to study alternate coordinate systems
in detail in Chapter 3. However, we will always need six independent scalars. The
remainder of this book describes methods for finding equations of motion—first for a
point (particle) and later for extended (rigid) bodies—and presents various techniques
for completely or partially solving them.
1.2 Organization of the Book
The next chapter reviews the physics of mechanics, covering Newton’s laws in depth.
We also start to solve simple problems. All the essential physical concepts that form
the foundation for the rest of the book are presented in that chapter. Our approach is
slightly unconventional in that we begin solving dynamics problems at the outset—in
Chapter 2—to highlight the meaning of Newton’s laws and how we incorporate the
underlying postulates7 into our methodology.
The remainder of the book is divided into five parts plus a set of four appendices.
We divide the book into parts to highlight the logical separation of main topics
and show how rigid-body motion builds on the key concepts of particle motion.
The material could be covered in one semester by leaving out certain topics or
stretched over multiple semesters or quarters. In Part One we restrict ourselves
to studying only the planar motion of single particles. Thus motion in only two
dimensions is studied; we thus need only four scalars to specify a particle’s state
7A
postulate is a basic assumption that is accepted without proof.
INTRODUCTION
7
rather than six. We do this to simplify the mathematics and focus on the key physical
concepts, allowing you to develop an understanding of the procedures used to solve
dynamics problems. You will solve an amazing array of real and important problems in
Part One.
Chapter 3 returns to first principles and lays out the mathematical framework for a
full vector treatment of kinematics and dynamics in the plane. Our focus is on the use
of various coordinate systems and approaches to treating velocity and acceleration.
Throughout the chapter we return to the same example: the simple pendulum. While
this example may seem a bit academic, our approach is to focus repeatedly on this
relatively simple system to emphasize the various new techniques presented and
explain how they interrelate and add value. At the end of the chapter these new
concepts are used to solve a selection of more difficult problems.
Chapters 4 and 5 present the concepts of momentum and energy, respectively, for a
particle. It is here that we begin to solve equations of motion for the characteristics of
trajectories (also called integrals of the motion). These ideas will be useful throughout
the remainder of your study of dynamics and form the foundation of modern physics.
Part Two presents an introduction to multiparticle systems (Chapters 6 and 7).
The previous concepts are generalized to simultaneously study many, possibly interacting, particles. In Chapter 6 we introduce two important examples of multiparticle
systems—collisions and variable-mass systems. Chapter 7 sets the stage for the rigidbody discussions in Parts Three and Four by analyzing angular momentum and energy
for many particles.
Part Three introduces rigid-body dynamics in the plane. We show (Chapters 8
and 9) how to specialize our tools to study a rigid collection of particles (i.e., particles
whose relative positions are fixed). In particular, the definition of equations of motion
is expanded to include the differential equations that describe the orientation of a
rigid body. We use these ideas to study a variety of important engineering systems.
We still confine our study to motion in the plane, however, to focus on the physical
concepts without being burdened by the complexity of three-dimensional kinematics.
It is here that we introduce the moment of inertia and, most importantly, the separation
of angular momentum.
Part Four develops the full three-dimensional equations that describe the motion of
multiparticle systems and rigid bodies. Part Four (Chapter 10) begins with the study
of the general orientation of reference frames, three-dimensional angular velocity,
and the full vector kinematics of particles and rigid bodies. Chapter 11 completes
the discussion by developing the equations of motion for three-dimensional rigidbody motion. It is here that we find the amazing and beautiful motion associated with
rotation and spin, such as the gyroscope and the bicycle wheel.
Part Five—Advanced Topics—allows for greater exploration of important ideas
and serves to whet the appetite for later courses in dynamics. Chapter 12 treats three
important problems in dynamics more deeply, exploring how the concepts in the book
are used to understand and synthesize engineered systems. This introduction is useful
for future coursework in dynamics and dynamical systems. Chapter 13 includes a
brief introduction to Lagrange’s method and Kane’s method. It serves as a bridge to
your later, more advanced classes in dynamics and provides a first look at alternative
techniques for finding equations of motion.
We have organized the book in a way that maximizes the use of problems and
examples to enhance learning. Throughout the text we solve specific examples—
sometimes repeated using different methods—to illustrate key concepts. Toward the
8
CHAPTER ONE
end of each chapter we include a tutorials section. Tutorials are slightly longer than
examples; they synthesize the material of the chapter and illustrate the important
ideas on real systems. The tutorials are an essential learning tool to introduce useful
techniques that may reappear later in the book. The tutorials vary widely in length,
depth, and difficulty. You may want to skim the longer or more difficult tutorials on the
first read and return later for reinforcement of key concepts or for practice on difficult
problems. We have intentionally incorporated this range of tutorials to maximize the
utility of the text for the widest possible audience and to make it a practical and helpful
reference throughout your career.8
We also include computation in many of our examples, tutorials, and problems.
Computation is central to modern engineering and an important skill to be learned.
It is integral to the learning and practice of dynamics. To simplify our presentation
and make it consistent throughout the book, we have exclusively used matlab for all
numerical work. There are many excellent numerical packages available (and some
students may want to code their own). We chose matlab because of its ubiquity,
its ease of use, and the transparent nature of its programming language. Our goal,
however, is not to teach the use of a particular programming tool but for you to
become comfortable with the full problem-solving process, from model building
through solution.
We end each chapter with a summary of key ideas, which contains a short list of the
main topics of the chapter. We intentionally minimize the prose in these sections to
make it as easy as possible to use for reference and review. Reading these sections does
not replace reading the chapters; they are meant only to serve as helpful references.
We used many sources in preparing this book and are indebted to a large number
of authors that preceded us. Our primary references are listed in the Bibliography.
In some cases, however, we highlight a particularly important result and direct you
to other references with more in-depth discussions or additional insights. Thus each
chapter has a Notes and Further Reading section, where we point out these sources.
Finally, we end each chapter in Parts One to Four with a problems section that
includes problems that address each of the topics of the chapter. We have tried to
provide problems of varying levels of difficulty and those that require computation.
We have not included problems sections in Part Five, as Chapters 12 and 13 are
intended as only an introduction to more advanced topics.
1.3 Key Ideas
.
.
A vector is a quantity with both magnitude and direction in space. The position of
point P relative to point O is the vector rP/O .
A reference frame provides the perspective for observations regarding the motion
of a system. A reference frame contains three orthogonal directions.
8 Because
Chapters 12 and 13 are similar to extended tutorials and are meant as only an introduction to
more advanced material, we do not include tutorials or problems in them.
INTRODUCTION
.
.
.
.
.
9
The velocity is the change in time of a position with respect to a particular reference
Id
frame. The velocity of point P relative to frame I is IvP/O =
dt (rP/O ).
A coordinate system is the set of scalars used to locate a point relative to another
point in a reference frame. Cartesian coordinates x, y, and z constitute the
most common coordinate system. We usually use (x, y, z)I to represent the Cartesian coordinates with respect to frame I . The rates of change (ẋ, ẏ, ż)I of the
Cartesian coordinates are called speeds.
The state of a particle consists of its position and velocity in a reference frame at
a given time.
The equations of motion are the three second-order differential equations for the
particle state whose solution provides the trajectory of a point.
An equilibrium point is a special solution of the equations of motion for which
the rates of change of all states are zero.
1.4 Notes and Further Reading
The modern formulation of dynamics is the culmination of more than two centuries
of development. For instance, while Newton presented the fundamental physics, the
concept of equations of motion and the formulation of the second law we know today
were given by Euler.9 The modern concept of a vector was introduced by Hamilton
in the mid-nineteenth century.10 A good, concise discussion of the early history of
dynamics can be found in Tenenbaum (2004). A more thorough treatment of the
history of mechanics is in Dugas (1988). We also recommend the book of essays
by Truesdell (1968) for insightful discussions of important historical developments.
Careful notation is essential for both learning dynamics and solving problems in
your professional career. Unfortunately, no universally accepted notation is in use.
In fact, there is much discussion among educators and practitioners over how to
balance simplicity and clarity. Our notation—particularly the use of reference frames
in derivatives—is closest to that of Kane (1978) and Kane and Levinson (1985).
A similar notational approach is used by Tenenbaum (2004) and Rao (2006). Our
notation for position is also used in Tongue and Sheppard (2005) with a variation
in Beer et al. (2007). Our qualitative definition of reference frames is similar to
that in Rao (2006). Other good discussions of the importance of reference frames
in dynamics can be found in Greenwood (1988), Kane and Levinson (1985), and
Tenenbaum (2004). Tenenbaum also has a similar and insightful discussion regarding
the distinction between coordinate systems and reference frames.
9 Leonhard Euler (1707–1783) was a Swiss mathematician and physicist. He is known for his seminal
contributions in mathematics, dynamics, optics, and astronomy. Much of our current notation is attributable
to Euler. He is probably best known for the identity eiπ + 1 = 0, often called the most beautiful equation
in mathematics.
10 Sir William Rowan Hamilton (1805–1865) was an Irish mathematician and physicist. He made fundamental contributions to dynamics and other related fields. His energy-based formulation is the foundation
of modern quantum mechanics.
10
CHAPTER ONE
1.5 Problems
1.1
What are the Cartesian coordinates of point P in frame I , as shown in
Figure 1.5?
I
ez
ey
2.27 m
1.92 m
P
3.64 m
4.71 m
O
1.23 m
ex
Figure 1.5 Problem 1.1.
1.2
Sketch and label the vectors rP/O , rP/Q, rQ/P in Figure 1.6.
I
P
O
Q
Figure 1.6 Problem 1.2.
1.3
Match each of the following definitions to the appropriate term below:
a. A perspective for observations regarding the motion of a system
b. A mathematical quantity with both magnitude and direction
c. Second-order differential equations whose solution is the trajectory of
a point
d. A set of scalars used to locate a point relative to another point
.
.
.
.
Vector
Reference frame
Coordinate system
Equations of motion
CHAPTER TWO
Newtonian Mechanics
In this chapter we reintroduce the physical principles that underlie the study of
dynamics. (We assume that you remember a bit from your physics classes.) In all that
we do, Newtonian methods are used for solving dynamics problems. That is, we will
solve problems using Newton’s three laws of motion to relate forces and acceleration.
This approach differs from the methods of Lagrange and Hamilton, which rely
on energy techniques. (A brief introduction to Lagrangian methods is provided in
Chapter 13.) To be sure, Lagrangian and Hamiltonian methods are important, are used
frequently, and may become a cornerstone of your subsequent dynamics education.
However, Newtonian methods have stood the test of time, are used regularly by
practicing engineers, and provide an important foundation for the study and practice
of dynamics. Using Newtonian methods, you can solve an amazing array of real
engineering problems—some of which are actually more difficult to solve with other
methods.
The purpose of this chapter is to instill a deep understanding of the physics of
motion, which is interchangeably called dynamics or mechanics. You will finish
this chapter with the basic skills needed for solving most dynamics problems. It is
interesting to note that almost all of the physics in the book is contained in this
chapter; no new laws of motion or other basic physical principles appear again
(with the exception of one assumption about rigid bodies discussed in Chapter 9).
In principle, the reader with great mathematical skills and insights could stop reading
at the end of this chapter and solve any dynamics problem. This is, of course, an
exaggeration, and we will provide many useful techniques and insights throughout
the book, but the fact remains that there is no new physics beyond Newton’s three
laws of motion.
2.1 Newton’s Laws
In 1687 Isaac Newton published the Principia, one of the greatest events—if not the
greatest—in the history of science. With this single book he overturned centuries of
12
CHAPTER TWO
misconceptions embedded in the Aristotelian idea that forces are necessary for bodies
to remain in motion.1 In the Principia, Newton provided not only a new philosophy
but also specific tools for solving real problems. (In particular, Newton developed
differential and integral calculus, though it is Leibniz’s notation that we use today.)
Here, then, are Newton’s three laws of motion as translated in 1729 by Andrew Motte:2
Law I Every body perseveres in its state of rest, or of uniform motion in a right line,
unless it is compelled to change that state by forces impressed thereon.
Law II The alteration of motion is ever proportional to the motive force impressed;
and is made in the direction of the right line in which that force is impressed.
Law III To every action there is always opposed an equal reaction; or the mutual
actions of two bodies upon each other are always equal, and directed to contrary
parts.
Newton’s first law is the explicit rejection of the Aristotelian idea that bodies
must be acted on by a force to remain in motion (even at constant velocity). Newton
elevated Galileo’s experimental observations to his first law of motion. This idea is
fundamental to modern dynamics and to the Newtonian method: forces cause the
motion of bodies to change.
Newton’s second law is the familiar statement that the rate of change of linear
momentum of a body—represented by the symbol p—is equal to the net force acting
on the body—represented by F.3 Newton’s second law is a vector relationship; hence
we use boldface letters for the force and linear momentum vectors.
The net force F is the sum, or resultant, of all force vectors acting on the body (see
Section 1.1).4 Newton recognized the not necessarily obvious fact that the net force
may produce an action in a direction different from any of the individual forces. (In
fact, this observation is Corollary I of his laws.) When solving a dynamics problem,
we begin by drawing each object and graphically illustrating every force vector acting
on it; the net force is just their geometric sum. You should remember this diagram
from statics; it is called a free-body diagram. Free-body diagrams are used throughout
our study of dynamics as well. A simple example of a free-body diagram is shown
in Figure 2.1. This vector nature of Newton’s second law is extremely important; we
explore it in depth in the next chapter.
Newton’s third law, which is often called the law of action and reaction,is probably
the most forgotten or misunderstood. Yet this law is of utmost importance to everyday
experience and engineered systems. Simply put, if an object A exerts a force on object
1 It was actually Galileo Galilei (1564–1642) who showed that,
in the absence of forces, bodies stay at rest
or in uniform motion, giving us the science of kinematics and the experimental method. However, Newton
placed the new science on a firm mathematical foundation and developed the tools for solving problems.
2 As with most scientific works of the day, the original was written in Latin.
3 Note that Newton’s second law actually only states proportionality, so to be absolutely correct we should
include an arbitrary constant in the equation relating force and the rate of change of linear momentum.
However, by convention, the units of force and linear momentum are selected to make this constant unity.
Thus, in the International System (SI) of units the newton is a derived unit equal to 1 kg-m/s2 .
4 For a brief review of vector algebra, consult Appendix B.
NEWTONIAN MECHANICS
13
FD
Fg
Figure 2.1 A simple example of a free-body diagram, consisting of a falling particle acted on
by two forces: gravity in the downward direction and aerodynamic drag resisting its fall in the
upward direction.
B, then object B exerts an equal and opposite force on object A. The recoil in a gun is
a common example. Another example is the force on our wrist when we hit a tennis
ball with a racket. Newton’s third law is extremely important for explaining flight: an
airfoil at a non-zero angle of attack pushes against the airstream as it moves forward;
by Newton’s third law, the air pushes back, generating lift.
Newton’s three laws—and the accompanying mathematical tools we will
develop—suffice to solve all problems posed in this book. We ignore dynamics at
the atomic scale, where Newtonian physics breaks down and quantum mechanics
takes over. We also ignore motion at speeds near the speed of light, where special
relativity must be taken into account.
Although a complete vector treatment of Newton’s second law is premature, we
can still solve meaningful problems with only a scalar representation. Recall from
Chapter 1 our discussion of position and velocity as described by three scalars and
their derivatives. Newton’s second law applies independently to each of these scalar
coordinates. For example, in the x-direction, we write
fx = ṗx = mẍ,
mẋ is the linear momentum in
where fx is the sum of all forces in the x-direction, px =
the x-direction, and ẍ is the second derivative of x with respect to time (acceleration).
(Appendix A supplies a brief review of some essential results from calculus if you
are rusty.)
Example 2.1 Straight-Line Motion with No Force
The simplest possible system in dynamics is one in which there are no net forces, that
is, fx = 0. In this case Newton’s second law reduces to the simple equation of motion
ẍ = 0.
(2.1)
To solve the equation of motion for x(t) and ẋ(t), we use the definition ẍ =
Multiplying Eq. (2.1) by dt and integrating yields
d ẋ
dt .
d ẋ = ẋ(t) + C1 = 0,
(2.2)
14
CHAPTER TWO
where C1 is an integration constant. Multiplying both sides of Eq. (2.2) by dt and
integrating a second time yields
dx + C1dt = x(t) + C1t + C2 = 0,
(2.3)
where C2 is another integration constant. Setting t = 0 in Eqs. (2.2) and (2.3), we
find that the constants are C1 = −ẋ(0) and C2 = −x(0), where x(0) and ẋ(0) are the
initial conditions. The solution is
ẋ(t) = ẋ(0)
x(t) = x(0) + ẋ(0)t.
This example is a simple, vivid demonstration of our problem-solving methodology. We use Newton’s second law to find an equation of motion (in this case, Eq. (2.1))
and then solve the equation of motion to find a trajectory in terms of initial conditions. In this case, the equation of motion was particularly easy to solve using direct
integration (unfortunately, that is unusual!).
Example 2.2 Straight-Line Motion with Constant Force
In this example we consider fx to be a constant force, in which case the equation of
motion is5
ẍ =
fx
.
m
(2.4)
Integrating Eq. (2.4) as in Example 2.1, we obtain
f
d ẋ = ẋ(t) + C1 = x t
m
(2.5)
and
dx +
C1dt = x(t) + C1t + C2 =
fx 2
t .
2m
(2.6)
Setting t = 0 in Eqs. (2.5) and (2.6), we find that the constants are C1 = −ẋ(0) and
C2 = −x(0). The solution trajectory is
ẋ(t) = ẋ(0) +
fx
t
m
x(t) = x(0) + ẋ(0)t +
fx 2
t .
2m
You may recognize the equation for x(t) from introductory physics; it describes the
displacement of a particle undergoing constant acceleration fx /m.
5 Note that when writing equations of motion, we solve for the second-order variables (e.g., ẍ), which are
the unknowns in the sense of a system of algebraic equations.
NEWTONIAN MECHANICS
15
Example 2.3 Straight-Line Motion with Position-Dependent Force
Consider a force fx (x) that is a function of position x. Thus Eq. (2.4) becomes
ẍ =
fx (x)
= a(x).
m
(2.7)
The quantity fx /m = a(x) has units of acceleration and is often referred to as the
acceleration of the mass. You can also view it as the specific force acting on mass m.6
Eq. (2.7) is a separable differential equation (see Appendix C). To integrate it, first
d ẋ
multiply both sides by dx = ẋdt and then use the definition ẍ =
dt on the left side.
Integrating from time t1 to time t2 yields
x(t2)
t2
t2
1
1
a(x)dx.
(2.8)
ẍ ẋdt =
ẋd ẋ = (ẋ(t2))2 − (ẋ(t1))2 =
2
2
t1
t1
x(t1)
Replacing ẋ(ti ) by the velocity v(ti ), for i = 1, 2, allows us to write Eq. (2.8) a bit
more compactly:
x(t2)
a(x)dx.
(2.9)
v 2(t2) = v 2(t1) + 2
x(t1)
Eq. (2.9) is an elegant and compact expression that relates the velocity directly to
position. This expression can be helpful and convenient for some problems, particularly when the acceleration (specific force) is an integrable function of position. For
instance, consider the case of constant acceleration. In this case, Eq. (2.9) yields an
alternative velocity equation:
v 2(t2) = v 2(t1) + 2a(x(t2) − x(t1)).
In addition to forces that depend on position, we often encounter forces that depend
on velocity (see Tutorial 2.2).
2.2 A Deeper Look at Newton’s Second Law
It is helpful to pause for a moment to contemplate the significance and, more importantly, the assumptions underlying Newton’s laws. This exercise is more than mere
pedagogy. Newton postulated important universal facts that form the foundation for
his laws and that will inform all we do. Understanding these postulates is essential
for understanding dynamics and solving problems.
For instance, why are Newton’s laws called laws and not theories? Why is it not
Newton’s Theory of Motion? A useful explanation appears in a publication by the
National Academy of Sciences.
Definition 2.1 A law is a descriptive generalization about how some aspect of the
natural world behaves under stated circumstances.
6 In
this context, the word specific means divided by a quantity representing an amount of material. The
specific force is the force divided by mass.
16
CHAPTER TWO
Contrast this definition to the definition of a theory, the most important endpoint
of a scientific endeavor.7
Definition 2.2 In science, a theory is a well-substantiated explanation of some
aspect of the natural world that can incorporate facts, laws, inferences, and
tested hypotheses.
Newton’s three laws—most importantly, his second law—provide a predictive
description of the behavior of objects subjected to forces. They allow us to analyze
our observed universe (the motion of the planets being the most obvious—and first—
successful application of his laws) and to synthesize engineered devices. They do not,
however, explain why or how objects behave the way they do. In particular, Newton
did not explain what a force is or how the concept of force arises from first principles.
In fact, this omission is one of the greatest objections to Newton’s Principia. Newton
did not explain mass or inertia. Nor did he explain the meaning of acceleration or
the absolute space relative to which acceleration is measured. All these ideas are
important for understanding how to use Newton’s second law—and its limitations—
so we explore them in more detail in the following subsections.
2.2.1 The Concept of Force
The concept of force in Newton’s second law is the most ill-defined and philosophically difficult concept in classical mechanics. What, in fact, did Newton mean by
“force”? What is a force? How does it arise? As pointed out earlier, Newtonian mechanics consists of laws, not theories. In fact, there is no accepted explanation for
the concept of a force. From almost the moment the Principia was published, scientists and philosophers criticized the concept of force as devoid of meaning. Modern
physicists have completely eliminated the concept of force from the formulation of
almost all statements of the laws of physics (such as quantum mechanics and general relativity). In fact, in Chapter 13 we will discuss how classical mechanics can be
reformulated to avoid the use of forces entirely (at least most of the time).
Nevertheless, Newtonian mechanics was—and remains—profoundly successful
and forms the foundation of what we will study in this book. We need only accept
that a force is simply an abstract concept that causes objects to behave in a predictable
way. A remarkable array of problems can be solved using Newtonian methods. We will
introduce a variety of forces without worrying about a precise physical explanation.
By using forces in Newton’s second law, we can describe and predict the macroscopic
motion of objects. This approach is a compact and elegant tool for engineering design
and analysis.
2.2.2 The Concept of a Point Mass
Newton’s use of the term “body” in his statement of the three laws is a bit misleading.
In fact, his laws apply to one and only one sort of object: the point mass. What is a
point mass? Unfortunately, there is no good explanation in Newton’s laws. (They are
7 For
example, Darwin’s Theory of Evolution or Einstein’s Theory of Relativity.
NEWTONIAN MECHANICS
17
laws, not theories!) In fact, there is no good explanation for the concept of mass—
Newton defined it as the quantity of matter. For our purposes, it does not matter.
A point mass is an infinitesimally small body, or particle, of mass m that behaves as
predicted by Newton’s second law. In fact, we measure mass by observing a particle’s
behavior under Newton’s second law. We use the term “particle” interchangeably with
the term “point mass.”
An extended body is modeled as a collection of point masses. However, throughout
Part One of the book we often treat an extended body as if it were a single point mass.
This approximation turns out to be fine, though the rigorous proof is not given until
Chapter 6. Part Three shows how to treat the motion of extended bodies by examining
Newton’s second law for each of the (possibly infinite) constituent point masses.
The main idea is that Newton’s laws only apply to point masses. Remembering
this point can help you avoid many pitfalls. To make it easier to remember, we embed
our notation with little reminders. For example, we assign to every point mass a label,
such as P or Q. The net force acting on particle P is written as FP . For clarity, we
also add the subscript P to the particle mass mP . For a collection of N particles, we
use an index, such as i = 1, . . . , N , to label each particle and mi to denote its mass. In
general, to find the equations of motion of the collection, Newton’s laws are applied
to each particle separately. Part Three shows how to simplify this procedure when
the particles in the collection have fixed relative positions (that is, when they form a
rigid body).
2.2.3 Acceleration and Absolute Space
Our statement of Newton’s second law emphasized that the rate of change of momentum is relative to a reference frame I . However, we failed to indicate what this
reference frame is or why it matters. There is an inherent assumption in Newton’s
second law: the second law applies only when the acceleration is relative to a special
frame of reference. Newton called this special frame absolute space. We today call it
the inertial frame. We often use the letter I to label an inertial reference frame.
Unfortunately, there is no good definition of absolute space, which is another
weakness of Newton’s laws. It is common to refer to the “fixed stars” as absolute space.
But, of course, the stars are not fixed. One of Einstein’s great accomplishments in his
development of the Theory of General Relativity was the Equivalence Principle. This
principle removed the need for absolute space from Newton’s laws. Einstein showed
that a frame of reference falling freely in a gravitational field is an inertial frame,
which implies Newton’s laws hold in such a frame (e.g., the interior of a space station
in orbit).
For our purposes, the inertial frame is an essential abstraction that need not be explained physically. It usually suffices to choose a reference frame whose acceleration
is relatively small compared to the accelerations of the particles of interest. In this
case, any errors introduced are negligible. (We will give mathematical substance to
this approximation in Chapter 3.)
Nonetheless, the concept of an inertial frame is essential. The first thing we do in
every dynamics problem is draw the inertial frame to remind ourselves that the laws
of motion apply only in this frame. As in Chapter 1, we specify the (inertial) frame to
which the velocity is referred. The same notation is used for the acceleration vector
18
CHAPTER TWO
Ia
P/O ,
which is the instantaneous rate of change in time with respect to the inertial
Id I
frame of the velocity, that is, IaP/O =
dt vP/O . Using Cartesian coordinates with
d ẋ
d ẏ
d ż
respect to frame I , the acceleration is (ẍ, ÿ, z̈)I , where ẍ =
dt , ÿ = dt , and z̈ = dt .
It is important to remember the concept of Newtonian relativity.This concept states
that any reference frame moving at constant velocity but not rotating relative to an
inertial frame is also an inertial frame. This can be proven rather succinctly using the
ideas and notation of Chapter 3 (see Section 3.6).
2.2.4 Anatomy of Newton’s Second Law
At this point we are in a position to write Newton’s second law in the form to be used
mP IvP/O denote the linear momentum vector of
throughout the book. Let IpP/O =
particle P with mass mP . Newton’s second law states the time derivative with respect
to the inertial frame of the linear momentum of a point mass P is equal to the total
force FP acting on P , that is,
FP =
Id dt
pP/O = mP IaP/O .
I
(2.10)
Figure 2.2 graphically identifies the various notational elements in this representation
of Newton’s second law.
Note that, to arrive at the form of Newton’s second law in which force equals mass
I times acceleration, we factored the mass out of the derivative dtd IpP/O . It is not
correct to add a term ṁP IvP/O , which arises from using the product rule to evaluate
the derivative. Recall again that Newton’s second law applies only to point masses. If
a point mass has no extent, it can’t gain or lose mass. Thus in Newtonian mechanics,
ṁP = 0. Including a mass derivative term is okay only at relativistic speeds where the
Vector derivative The right-hand side
of Newton’s second law is the derivative
of the linear momentum with respect to
an inertial frame
Force The sum of all
the force vectors acting
on particle P
Linear momentum
of particle P
I
d
FP = — (IpP/O)
dt
Point mass Newton’s
second law only applies
to a point mass
I
= mP aP/O
Inertial frame Newton’s
second law only applies to
the acceleration with respect
to an inertial frame
Mass of particle P
Acceleration
of particle P
Origin of inertial frame
Figure 2.2 Anatomy of Newton’s second law.
NEWTONIAN MECHANICS
19
mass changes, and then all sorts of other problems arise. (If you are wondering how
rockets fly by expelling mass, we cover that in detail in Chapter 6.)
2.2.5 Conservation Laws
You may have noticed that Newton’s first and second laws are intimately related.
Letting the force in Eq. (2.10) equal zero and integrating with respect to time results
in a simple restatement of Newton’s first law:
I
pP/O = constant.
(2.11)
In words, Eq. (2.11) states that when the total force on a particle is zero, the linear
momentum of the particle is a constant of the motion. This is just Newton’s first law!
We often say that the momentum is conserved. The concept of conserved quantities
is an important one in dynamics and we return to it often in the book. In fact, it is
useful to carefully define it here.
Definition 2.3 A scalar or vector function of the state of a particle or multiparticle
system is conserved if it remains constant throughout the trajectory of the
particle or system.
We introduce a conservation law when a quantity is conserved under some general
set of circumstances. In fact, Eq. (2.11) already states our first conservation law:
conservation of linear momentum of a particle. Although this law is just a restatement
of Newton’s first law, it is useful to start our practice of highlighting important
conservation laws here.
Law 2.1 The law of conservation of linear momentum of a particle states that,
when the total force acting on a particle is zero, the linear momentum of the
particle is a constant of the motion.
Conservation laws can be incredibly useful—they may reduce the complexity of
the equations of motion, provide important constraints on the trajectory, or provide
tools for checking the accuracy of our analysis and simulations. Don’t forget the
constraints or conditions required to invoke the conservation law. When we find a
conservation law by integrating the equations of motion once, as in this case, we call
the resulting conserved quantity a first integral of the motion.
2.3 Building Models and the Free-Body Diagram
Since Newton’s second law only applies to a point mass, it is sensible to ask how
to use it to solve for the motion of more complex objects. In engineering practice
we may want to understand how cars move, airplanes fly, or submarines maneuver.
These objects seem rather different from point masses. The art of dynamics comes in
building representative models for non-point masses out of elements we understand
and can use to find equations of motion. In this book, we derive trajectories for point
masses, collections of point masses, and rigid bodies of basic shapes (rods, disks,
spheres, etc.). Where practicable we show that finding the equations of motion for
20
CHAPTER TWO
these simple elements is equivalent to finding the equations of motion for the original
system.
The first step of any dynamics problem is to formulate a model—using basic modeling elements and various connective abstractions—for the system being studied.
The challenge is to find the simplest model that provides meaningful results. Simplicity is essential. To achieve it, we often render some model components as massless.
However, oversimplification can lead to trouble if you are not careful. For example,
a massless component does not satisfy Newton’s third law!
Once you have modeled a system, the reference frames, coordinates, and forces
need to be identified. Only then can Newton’s second law be used to find the equations
of motion and trajectories. We start by drawing for each mass a free-body diagram
that explicitly identifies the relevant force vectors. You may be familiar with freebody diagrams from statics or physics. In statics, you vectorially add all the forces in
the free-body diagram and set the sum equal to zero in each orthogonal direction—as
required for a static equilibrium. In dynamics, the vector sum of the forces is proportional to the acceleration. In every problem you solve, draw a free-body diagram
before writing down Newton’s second law. This habit is essential for solving dynamics
problems. It should soon become second nature.
Figure 2.3 depicts a physical system and a suitable model constructed from point
masses. The Watt flyball governor, shown in Figure 2.3a, was one of the first examples
of a feedback control system. Widely used in the control of mills during the seventeenth century, the first steam governor was designed by James Watt in 1788. The
governor is a beautiful example of using simple dynamics in design. The actual governor pictured in Figure 2.3a is reasonably complicated, with many gears, masses,
and linkages. However, it can be reduced to a very simple model using only point
masses and massless rods, as shown in Figure 2.3b. The free-body diagram corresponding to this model is shown in Figure 2.3c. Even this bare-bones model produces
a remarkable richness of motion and displays the fundamental principles underlying
the flyball governor. We explore it in more depth in Chapter 10.
T1
θ
l
y
m
l
M
Ω
(a)
T2
(b)
mg
g
T2
T2
Mg
(c)
Figure 2.3 The Watt flyball governor, designed in 1788, was used to control the rotational
speed of a steam engine drive shaft. Though a complex mechanical system, it is relatively simple
to model using three point masses and four massless rods. (a) Flyball governor. (b) Governor
model. (c) Free-body diagram.
NEWTONIAN MECHANICS
21
2.4 Constraints and Degrees of Freedom
In Examples 2.1 and 2.2 we treated motion in one dimension only. A dynamicist
would say that the particle had only a single degree of freedom. As discussed in
Chapter 1, a particle trajectory is defined by three positions and three speeds. In the
three-dimensional world, particles have three degrees of freedom: they are free to
move in three orthogonal directions. However, not every problem we solve will consist
of particles free to move in every direction. A particle sliding on top of a table, for
example, or a hockey puck sliding on a flat ice rink, as in Figure 2.4, moves in only
two directions. Such a particle has two degrees of freedom.
We use the idea of degrees of freedom often in dynamics to gain understanding
of a problem and to avoid unnecessary work. In other words, if we know from the
beginning that a particle has fewer than three degrees of freedom, then we should need
fewer equations of motion. We associate a single scalar coordinate with each degree
of freedom and produce a single equation of motion for each coordinate.8 Thus, a two
degree-of-freedom problem will have only two nontrivial equations of motion. We
call out this important point in the following definition.
Definition 2.4 The number of degrees of freedom of a collection of particles is the
number of independent coordinates needed to describe the position of every
particle. For a collection of rigid bodies, the number of degrees of freedom is
equal to the number of independent coordinates needed to describe the position
and orientation of every rigid body.
If there are more equations of motion than degrees of freedom, we have to simultaneously solve the constraint equations. Each constraint equation represents a
reduction of one degree of freedom. Constraints can be either explicit or implied. Either way, the presence of a constraint in a dynamics problem implies that there exists
a coordinate that does not play a role in the solution. For instance, for the hockey
puck in Figure 2.4, we can use Cartesian coordinates to track its position. If x and
y describe the horizontal position, then a mathematical statement of the constraint is
z = 0. Examples 2.1 and 2.2 are one-degree-of-freedom problems—in each example,
there are implied constraints that prevent motion in the other two directions. Throughout Parts One, Two, and Three we treat only one- and two-dimensional systems. That
is, we assume an implied constraint z = 0 and sometimes y = 0 as well.
For simple examples like the hockey puck, labeling the degrees of freedom and
constraints may seem to be unnecessary. However, understanding degrees of freedom
and constraints is an important part of understanding dynamics. Not all constraints are
so simple, and the degrees of freedom do not always align nicely with simple Cartesian
coordinates. For instance, if a particle is constrained to move on a curved surface,
then there is still one constraint equation—albeit more complicated than z = 0—
and
two degrees of freedom. For instance, a hemispherical table has the constraint
x 2 + y 2 + z2 = constant. When we treat multiple particles, the number of degrees
of freedom can become quite large—each particle nominally has three degrees of
freedom—or the number of constraints limiting the relative motion of the particles
can become large. And, to further complicate things, we may have constraints on the
8 Up
until now we have discussed only Cartesian coordinates—other coordinate systems are introduced in
Chapter 3.
22
CHAPTER TWO
Figure 2.4 A hockey puck sliding on the surface of an ice rink is an example of a two-degreeof-freedom system. Image courtesy of Shutterstock.
motion rather than on the configuration; for example, an ice skate can only move
along the direction of the blade, not perpendicular to it.9
Counting degrees of freedom is an important habit that ensures you have the correct
number of equations of motion. In certain problems—we will see some soon—the
constraint equations need to be used explicitly to obtain the same number of equations
of motion as there are degrees of freedom. In general, the number M of degrees of
freedom in a system of N particles is
M = 3N − K,
(2.12)
where K is the number of constraints. Note that the number of degrees of freedom in
a system of N rigid bodies and K constraints is M = 6N − K, since the orientation
of each body counts as three additional degrees of freedom.
In problems with reduced degrees of freedom, explicit constraints imply the presence of a force—called a constraint force. This observation results from a direct
application of Newton’s third law. Consider again a particle on a table. Assuming
that the force of gravity keeps the particle on the table, then the table must apply
an equal and opposite force on the particle. In this case, the force of constraint is
a normal force—a force orthogonal to the direction of motion. Usually, we are not
interested in the constraint forces and seek to eliminate them from the equations of
motion. Sometimes we may solve for these forces in terms of the particle trajectory
to be certain our system is engineered correctly. (We do not want the table to break
under the weight of the particle.) These ideas are demonstrated in a classic example
at the beginning of Chapter 3.
9 Motion
constraints are often called nonholonomic constraints. We discuss these in Chapter 13.
NEWTONIAN MECHANICS
23
P
I y
s
x
(a)
(b)
Figure 2.5 A roller coaster and the corresponding model. Image (a) courtesy of Shutterstock.
Example 2.4 Modeling a Roller Coaster
Suppose we want to analyze the dynamics of a roller coaster like the one shown in
Figure 2.5a. The simplest model would confine the tracks to a horizontal or vertical
plane and model the car as a single point mass, as shown in Figure 2.5b. How many
degrees of freedom does this model possess?
The roller-coaster car’s position can be specified using Cartesian coordinates
(x, y)I . However, we know that the car must remain on the track. Suppose the track
shape can be defined by a function y = f (x). This is then a constraint on the geometry,
reducing the number of degrees of freedom from two to one. We can solve for
the motion of this single-degree-of-freedom model by finding equations for (x, y)I
subject to the constraint y = f (x) or, alternatively, we can use a single coordinate,
s, to represent the distance the car has traveled along the track. Remember, since the
model has one degree of freedom, we only need one scalar coordinate. We call s a
path coordinate. Path coordinates are discussed at some length in Chapter 3.
Example 2.5 Modeling a Three-Link Robot Arm
Figure 2.6a is a picture of a common three-link robot arm used to move and place
objects. It consists of a shoulder joint, upper arm, elbow, lower arm, wrist joint, and
hand. Suppose we wish to model the arm and develop equations describing its motion
(presumably to develop a control system for it). The goal would be to determine the
position of the hand. We discuss this problem in more detail in Chapter 8. Here,
however, we are interested in modeling the system, determining how many degrees
of freedom it has, and choosing appropriate coordinates to describe the system.
Figure 2.6b shows our model of the robot arm consisting of three linked rigid
bodies (the upper and lower arms and hand). Each joint can rotate about only a single
axis perpendicular to the plane of motion. We can thus define the angle of each joint
by a single coordinate. By inspection, then, this system has three degrees of freedom.
Thus three scalar coordinates suffice to describe the state of the system. In Figure 2.6b
we use the three angular coordinates (θ1, θ2, θ3) to describe the angles of each joint
relative to the line through the previous arm. However, as mentioned earlier, the choice
of scalar coordinates in a problem is not unique. For instance, in this problem we could
equally well choose as coordinates the angles of each joint relative to the inertial
x-axis. Although the motion of the hand would be the same (it would follow the
24
CHAPTER TWO
I
θ2
θ3
O
(a)
θ1
(b)
Figure 2.6 A three-link industrial robot arm and the corresponding model. Image (a) courtesy
of Shutterstock.
same trajectory x(t) and y(t)), how we describe it using scalar coordinates would be
different. Part of the art of dynamics is finding the best set of scalar coordinates for a
problem.
2.5 A Discussion of Units
We end this chapter with a brief discussion of units. For all problems in this book we
use SI units.10 This set of units is the internationally accepted system that uses metric
measures of length and mass. All SI units and subunits in a category (length, say) are
related by factors of 10, and the power of ten is identified explicitly by the prefix.
Thus a kilometer is equal to 1,000 meters. We supply a list of the common prefixes
in Table 2.1.
The SI system is an absolute system. It is based on three standards: length (the
meter (m)), mass (the kilogram (kg)), and time (the second (s)). Each of these fundamental units can be related to a well-defined physical phenomenon and is independent
of the location of the measurement (hence the term “absolute”). No matter where we
solve problems or take measurements, the meter is the meter, the second is the second,
and the kilogram is the kilogram. In case you were wondering, the second is defined
to be the duration of 9,192,631,770 periods of the electromagnetic wave emanated
during the transition between the two hyperfine energy levels of the ground state of
the cesium 133 atom. The meter is the length of the path traveled by light in vacuum
during a time interval of 1/299,792,458 of a second.11
The unit of force in SI—the newton—is a derived unit: 1 newton is equal to 1 kgm/s2. In words, under 1N of force, a mass of 1kg will accelerate at a rate of 1m/s2. The
10 The common abbreviation for the system, SI, comes from the French: Le Système International d’Unités.
11 The kilogram is less well defined;
it is the only base unit still defined by a physical object. The kilogram
is defined as the mass of a platinum-iridium bar being kept at the Bureau International des Poids et Mesures
near Paris. Unfortunately, the kilogram was recently discovered to have lost 50 micrograms over the past
100 years for unknown reasons. Efforts are under way to develop an atomic standard for the kilogram.
NEWTONIAN MECHANICS
25
Table 2.1
The common International System (SI) prefixes and conversions between
U.S. Customary (USC) and SI units
Multiple
Symbol
Prefix
SI
USC
109
G
giga
1m
3.28 ft
106
M
mega
1 km
0.621 mi
103
k
kilo
2.54 cm
1 in
mm3
6.10 × 10−5 in3
10
da
deca
1
10−1
d
deci
1 m3
35.3 ft3
10−2
c
centi
1 m/s
3.28 ft/s
10−3
m
milli
1 km/h
0.621 mi/h
m/s2
10−6
μ
micro
1
10−9
n
nano
1 kg
0.0685 slug
10−12
p
pico
1N
0.225 lb
1J
0.738 ft-lb
1W
1.34 × 10−3 hp
1 N/m2 (Pa)
1.45 × 10−4 lb/in2 (psi)
3.28 ft/s2
weight of an object is a force equal to the mass of an object times the acceleration of
gravity g at sea level (g ≈ 9.81 m/s2). Weight is thus equivalent to the force of gravity.
A 1 kg mass weighs 9.81 N.
Despite governmental efforts to move the United States to the SI system, many
practicing engineers still use the U.S. Customary (USC) system of units. It is thus
helpful to be familiar with USC units, though we avoid using them in this book.
In the USC system, the fundamental standards are length (the foot (ft)), time (the
second (s)), and force (the pound (lb) or pound-force (lbf )). The second is the same
as the SI second (the only common unit between the two systems). The foot is equal
to 0.3048 m. The pound is equal to the sea level weight of a standard pound located
at the National Institute of Standards and Technology in Boulder, Colorado. Because
force, rather than mass, is the fundamental standard, USC units are not an absolute
system. The definition of a pound depends on location—in this case, the mean sea
level. The unit of mass (the slug) is thus a derived unit. To form a set of units consistent
with Newton’s second law, 1 slug is thus defined as the mass that accelerates 1 ft/s2
when acted on by 1 lb of force (or 1 lbf ). The slug is thus a derived unit equivalent to
1 lb-s2/ft. Thus the standard pound has a mass of 1/32.2 slugs.
2.6 Tutorials
Even without the notational tools of the next chapter we can solve interesting and
important problems. As in earlier examples, in this section we study problems in one
dimension, where Newton’s second law simplifies to the scalar equation fx = mẍ
26
CHAPTER TWO
(or fy = mÿ), where fx (or fy ) refers to the total force on the point mass in the xdirection (or y-direction). In other words, we treat only problems with one degree of
freedom here, except for the final problem, which examines a simple two-dimensional
(two-degree-of-freedom) problem as an introduction to the next chapter. Many of the
concepts and solutions presented here pop up again and again throughout the book.
You should finish this section with an understanding of what our objectives are in all
dynamics problems, the process we go through to find equations of motion, and even
how we solve those equations (when possible) to develop insight into the motion of a
body. The techniques introduced here are the same as those used for more complicated
problems later in the book. For example, we always start by drawing reference frames
and a free-body diagram.
The tutorials are sorted by the type of force acting on the point mass. It may
surprise you, but there are only a few types of force encountered in the book (and
in practice). We try to cover most of them here, albeit in only one dimension. Forces
generally fall into one of two categories: contact forces and field forces. As the name
implies, contact forces arise from the contact between two objects (or an object and
a stationary surface). We often call a contact force perpendicular to the direction of
motion a normal force. A contact force in the direction of motion is a tangential
force. The main tangential contact forces we will be studying are friction forces (see
Tutorials 2.1 and 2.2). Another example of a tangential contact force is aerodynamic
drag (see Tutorial 2.5). The main normal force encountered is the constraint force.
We discuss it at length in the next chapter.
Field forces act on a body at a distance. Three examples of a field force are gravity,
electrostatics, and magnetism. In this book, the primary field force encountered is
gravity. For objects that remain close to a fixed distance from the gravitating center
(the center of the earth, say), we approximate the acceleration due to gravity as
constant, represented by g. For objects farther away from the gravitational center,
we use Newton’s inverse-square law of gravity (see Tutorial 2.5).
Tutorial 2.1 Sliding Friction
Consider a (point) mass m sliding on a horizontal surface, where x denotes the position
coordinate, as depicted in Figure 2.7a, and ẋ is the particle speed.12 Let N denote the
magnitude of the normal force exerted by the table on the particle in the vertical
direction. One simple model of the interaction force between the surface and the
particle is called kinetic or Coulomb friction.13 Assuming ẋ > 0, we have
fx = −μc N,
where μc is the (unitless) coefficient of Coulomb friction. The minus sign indicates
that friction exerts a force in the opposite direction of the particle velocity, which is
to the right. Since the particle is not moving in the vertical direction, we can write
mÿ = 0 = N − mg.
12 Remember,
even if it looks like a box, we treat mass m as a particle!
13 Named for Charles-Augustin de Coulomb (1736–1806), the French physicist who developed the model.
NEWTONIAN MECHANICS
I
27
I
I
N
µcN
x
m
N
N
g
bx
x
m
mg
g
k(x – x0)
x
m
mg
mg
(a)
(b)
g
(c)
Figure 2.7 Free-body diagrams for (a) sliding friction, (b) viscous friction, and (c) a linear
spring acting on mass m.
The normal force, N , is an example of a constraint force. The constraint equation is
y = 0, which reduces the number of degrees of freedom from two to one. (There is
also the implicit constraint that the box motion is confined to the plane of the page.)
In the horizontal direction, the equation of motion is
ẍ =
fx
μN
= − c = −μc g.
m
m
As in previous examples with a constant force, the trajectory solution is easily
obtained by integration:
ẋ(t) = ẋ(0) − μc gt
1
x(t) = x(0) + ẋ(0)t − μc gt 2.
2
Note that this model is valid only as long as the particle is sliding to the right. In
the case where the particle is not moving, that is, ẋ = 0, we can use the static friction
model. The coefficient of static friction μs is typically greater than or equal to μc .
The static friction model is valid only when the particle is not moving relative to the
surface—in which case there is no equation of motion. In other words, an applied
force must exceed the static friction for motion to occur. Once the particle is moving,
the kinetic friction model applies.
Tutorial 2.2 Viscous Friction
Another type of friction, known as viscous friction, is proportional to the speed of the
particle and is directed opposite the particle velocity. Consider a sliding (point) mass
m attached to a damper, as shown in Figure 2.7b. A damper is a device that reduces
oscillations or vibrations using viscous friction (e.g., an automobile shock absorber).
In this system, the force from the damper on the mass depends on the velocity:
fx = −bẋ,
28
CHAPTER TWO
and the equations of motion are
b
ẋ
m
mÿ = N − mg.
ẍ = −
Note that we also have the constraint that y = ẏ = ÿ = 0; that is, we assume the
motion is one dimensional. This assumption yields the expression N = mg for the
normal force. (The normal force is an example of a constraint force.) The equation of
motion in the x-direction can be written as a first order, separable differential equation
(see Appendix C) using the change of coordinates z = ẋ. We have
ż =
dz
b
= − z.
dt
m
(2.13)
Separating variables in Eq. (2.13) and then integrating the result yields
dz
b
dz
b
= − dt which implies
= − t.
z
m
z
m
The solution is
ln z + C1 = −
b
t,
m
(2.14)
where C1 = − ln z(0). Plugging C1 into Eq. (2.14) and taking the exponential of both
sides gives
z(t) = z(0)e− m t
b
or, in the original coordinates,
ẋ(t) = ẋ(0)e− m t .
b
(2.15)
Thus the speed of mass m subject to viscous friction decreases exponentially to zero.
Next we integrate Eq. (2.15) to find the position x(t):
b
mẋ(0) − b t
x(t) + C2 = ẋ(0)e− m t dt = −
e m ,
b
where C2 = −x(0) −
mẋ(0)
b .
The solution is
x(t) = x(0) +
b
mẋ(0) 1 − e− m t .
b
Taking the limit t → ∞, we see that the position of mass m asymptotically approaches
the final position x(0) + mẋ(0)
b .
Tutorial 2.3 Simple Harmonic Motion (The Ubiquitous Spring)
Consider a force that is a linear function of position:
fx = −k(x − x0),
(2.16)
NEWTONIAN MECHANICS
29
where k is a proportionality constant. Eq. (2.16) is a model of the force exerted on
a particle by a linear spring with spring constant k kg/s2 and unstretched length x0.
The spring-mass system is illustrated in Figure 2.7c. In this system, the equation of
motion
k
(2.17)
ẍ = − (x − x0)
m
is a separable differential equation (see Appendix C). Multiplying Eq. (2.17) by
dx = ẋdt and integrating yields
k
1
(x(t) − x0)2.
(2.18)
ẋd ẋ = (ẋ(t))2 + C1 = −
2m
2
Setting t = 0 in Eq. (2.18), we find that the constant C1 is a function of the initial
conditions and the spring parameters, that is,
C1 = −
Solving for ẋ(t) =
dx
dt
k
1
(x(0) − x0)2 − (ẋ(0))2.
2
2m
in Eq. (2.18), we obtain
dx
=
dt
−
k
(x − x0)2 − 2C1.
m
(2.19)
By separating the variables x and t in Eq. (2.19) and integrating, we obtain
dx
= dt = t + C2.
− mk (x − x0)2 − 2C1
In principle, we can use an integral table (or trigonometric substitution) to integrate
the left-hand side to obtain an implicit equation for the position x. Fortunately, there
is an easier approach.
First, make the change of coordinates z = x − x0 and let ω0 = mk . Note that
ż = ẋ and z̈ = ẍ, since x0 is constant. Under this change of coordinates, the equation
of motion Eq. (2.17) becomes
z̈ = −ω02z.
(2.20)
You may recognize Eq. (2.20) as the equation for simple harmonic motion with natural
frequency ω0. To solve Eq. (2.20), we assume a solution of the form
z(t) = A cos(ω0t) + B sin(ω0t),
so that
ż(t) = −Aω0 sin(ω0t) + Bω0 cos(ω0t)
and
z̈(t) = −Aω02 cos(ω0t) − Bω02 sin(ω0t) = −ω02z(t).
Thus our assumed solution is indeed correct.
(2.21)
30
CHAPTER TWO
5
x(t) (m)
x(t) (m/s)
0
–5
0.0
0.5
1.0
1.5
2.0
2.5
3.0
t (s)
Figure 2.8 Position and velocity responses of a mass-spring system with natural frequency
ω0 = 10 rad/s and rest length x0 = 1 m. The initial conditions are x(0) = 1.5 m and ẋ(0) = 0.
We can solve for the value of the constants A and B by setting t = 0 in the
expressions for z(t) and ż(t) to find that A = z(0) and B = ż(0)
ω0 . The solution in the
original coordinate x becomes
ẋ(0)
x(t) = x0 + (x(0) − x0) cos(ω0t) + ω sin(ω0t)
0
(2.22)
ẋ(t) = −(x(0) − x0)ω0 sin(ω0t) + ẋ(0) cos(ω0t).
(2.23)
The solution is plotted in Figure 2.8.
This result is extremely important. The solution has the form of a sinusoidal
oscillation about the unstretched length x0 with amplitude and phase given by the
initial conditions. Eq. (2.20) for simple harmonic motion occurs frequently and in
a variety of contexts other than linear springs. Often we can reduce the equations
of motion of much more complicated problems to Eq. (2.20), in which case we
immediately know the solution has the form of Eqs. (2.22) and (2.23).
We conclude with the common situation of a simple harmonic oscillator with a
constant driving force:
ẍ + ω02x =
F0
.
m
This has the same form as Eq. (2.17) if we identify the unstretched length with the
constant driving force:
x0 =
F0
.
k
Thus a simple harmonic oscillator with a constant driving force has the same solution
as shown in Figure 2.8, a sinusoidal oscillation about a mean offset given by F0/k.
NEWTONIAN MECHANICS
31
Tutorial 2.4 Damped Simple Harmonic Motion
This tutorial combines Tutorials 2.2 and 2.3. That is, we find the equation of motion
(and solution) for a mass connected to both a linear spring and a viscous damper. The
total force on the mass is thus
fx = −k(x − x0) − bẋ.
The resulting equation of motion from Newton’s second law is
ẍ = −
k
b
(x − x0) − ẋ.
m
m
This problem is most easily solved using the second approach from the previous
tutorial: guessing a solution. First we make the change of variables z = x − x0 and
the substitution ω0 = mk . We also make another common substitution by replacing b
with 2mζ ω0. The dimensionless parameter ζ is called the damping coefficient. These
substitutions yield the equation of motion
z̈ + 2ζ ω0ż + ω02z = 0.
(2.24)
Our approach to solving this system is to use two facts from Appendix C: (1) if
we have two solutions to a linear ordinary differential equation (ODE) then the sum
is also a solution, and (2) the solution of any linear ODE is an exponential. (The sine
and cosine solutions from the previous example can be written as a sum of complex
exponentials.) Since Eq. (2.24) is a second-order ODE, its solution must have the
form
z(t) = Az1(t) + Bz2(t),
where A and B are two arbitrary constants that are functions of the initial conditions.
Since Eq. (2.24) is linear, z1(t) must have the form
z1(t) = eλt ,
with
ż1(t) = λeλt
and
z̈1(t) = λ2eλt .
The variable λ is a function—as yet undetermined—of the physical parameters of the
problem. Substituting Az1(t) into Eq. (2.24) yields
Aeλt λ2 + 2ζ ω0λ + ω02 = 0.
(2.25)
Properly choosing λ to satisfy Eq. (2.25) will validate our guess. Then we find the
constants A and B in terms of the initial conditions by setting t = 0.
32
CHAPTER TWO
The only way for Eq. (2.25) to be true for all A and all time is if the quantity in
parentheses is zero:
λ2 + 2ζ ω0λ + ω02 = 0.
The solution to this quadratic equation is easily found from the quadratic formula.
We find two possible solutions:
λ1,2 = −ζ ω0 ± ω0 ζ 2 − 1.
We have thus determined λ for both of the guessed solutions, z1(t) and z2(t). For most
systems of interest, ζ < 1. Thus it is convenient and common practice to write the two
values of λ as complex numbers:
λ1,2 = −ζ ω0 ± iω0 1 − ζ 2.
The general solution becomes
√
√
2
2
z(t) = Ae −ζ ω0+iω0 1−ζ t + Be −ζ ω0−iω0 1−ζ t
√
√
2
2
= e−ζ ω0t Aeiω0t 1−ζ + Be−iω0t 1−ζ .
There are three types of damped oscillators of interest, depending on the value of ζ
(or equivalently, b). The most common case of interest is ζ < 1. In this case, we have
an underdamped oscillation of the mass about the rest position. Since the physical
motion of the spring must, of course, be real, the constants A and B are complex
conjugates. Using Euler’s equation on the complex exponentials, we can rewrite the
above general solution as a real equation to make it easier to work with:
z(t) = e−ζ ω0t (A + B) cos ωd t + i(A − B) sin ωd t ,
where A + B is purely real, A − B is purely imaginary (implying that i(A − B) is
real), and we made the substitution
ω0 1 − ζ 2 .
ωd =
The quantity ωd is called the damped frequency of oscillation. We can now replace
the complex constants A and B with the constants c1 = A + B and c2 = i(A − B) to
find the underdamped general solution:
z(t) = e−ζ ω0t c1 cos(ωd t) + c2 sin(ωd t) .
The constants c1 and c2 can be found in terms of the initial conditions. First set
t = 0 to obtain
c1 = z(0).
Differentiating the general solution and setting t = 0 yields
c2 =
ż(0) + z(0)ζ ω0
.
ω0 1 − ζ 2
This solution is plotted as a solid line in Figure 2.9.
NEWTONIAN MECHANICS
33
1.0
Underdamped (ζ = 0.2)
Overdamped (ζ = 3)
Critically damped (ζ = 1)
0.8
z(t) = x(t) – x0 (m)
0.6
0.4
0.2
0.0
–0.2
–0.4
–0.6
0.0
0.5
1.0
1.5
2.0
2.5
3.0
t (s)
Figure 2.9 Position response of a damped mass-spring system with natural frequency ω0 = 10
rad/s. The solutions for three different values of the damping coefficient are shown: ζ = 0.2
(underdamped), ζ = 1 (critically damped), and ζ = 3 (overdamped). The initial conditions are
z(0) = 1 and ż(0) = 0.
The second case of interest is ζ > 1. The solution to this case is called overdamped.
In the overdamped situation, there are no oscillations, and the mass position gradually
decays to the rest length of the spring. Mathematically, this occurs because λ is real.
The general solution for an overdamped system is
√
√
2
2
z(t) = Ae−ω0 ζ + ζ −1 t + Be−ω0 ζ − ζ −1 t .
To find the constants A and B, set t = 0 to obtain
A + B = z(0).
Differentiating the general solution and setting t = 0 in the result yields
B −A=
ż(0) + z(0)ω0ζ
.
ω0 ζ 2 − 1
After a bit of algebra, we find
ż(0) + z(0)ω0 ζ − ζ 2 − 1
A=−
2ω0 ζ 2 − 1
ż(0) + z(0)ω0 ζ + ζ 2 − 1
B=
.
2ω0 ζ 2 − 1
The overdamped solution is plotted as a dashed line in Figure 2.9.
For ζ = 1 the system is called critically damped. In the critically damped case,
λ1 = λ2 = −ζ ω0. Like the overdamped response, the critically damped response has
34
CHAPTER TWO
I
I
cy2
x
y
y
m
m
GMm
—
(Re + y)2
mg
(a)
(b)
Figure 2.10 Free-body diagrams for (a) particle falling under gravity and (b) the tossed point
mass.
no oscillation; it decays directly to the rest length. In fact, critical damping is the
boundary between the damped oscillatory response and the overdamped response.
We leave it as a problem to show that the general solution to the critically damped
system is
z(t) = z(0)e−ζ ω0t + (ż(0) + z(0)ζ ω0)te−ζ ω0t .
(2.26)
The critically damped solution is plotted as a dash-dot line in Figure 2.9.
Tutorial 2.5 Particle Falling under Gravity with Air Resistance
This tutorial treats a third type of friction, which in this case is proportional to speed
squared, and introduces a more exact expression for gravity. In addition to formulating
his laws of motion, Newton also postulated a universal law of gravity. In this law, the
force of gravity between two masses is an inverse square law: the magnitude of the
force is inversely proportional to the square of the distance between the two masses.
The direction of the force is along the line connecting the two masses and acts to
pull them together. Let y denote the height of a particle of mass m, as shown in
Figure 2.10a; ẏ denotes speed, which is negative because the particle is falling.
Mass m is falling toward a larger mass M. Assume that the second (point) mass
is the earth, so M = 5.9742 × 1024 kg. The distance between m and M is r =
Re + y, where Re = 6,378,100 m is the earth’s radius. In SI units the gravitational
proportionality constant is G = 6.673 × 10−11 m3kg−1s−2. The total force on mass
m is the sum of the force of gravity (which pulls the particle downward) and the force
due to air resistance. We assume that the force of air resistance acts in the opposite
direction to the particle velocity and is proportional to speed squared. The force acting
on mass m is
fy = −
GMm
+ cẏ 2,
(Re + y)2
where c is the coefficient of air resistance with units kg/m.14 Using Newton’s second
law, we obtain the equation of motion:
= (1/2)ρCD A, where CD is the dimensionless drag coefficient, A is the area of the “point” mass in
contact with the air flow, and ρ is the air density.
14 c
NEWTONIAN MECHANICS
35
ÿ = −
GM
c
+ ẏ 2.
2
(Re + y)
m
(2.27)
Eq. (2.27) looks a bit tricky to integrate by hand, so we proceed by integrating
numerically using the matlab function ODE45. (See Appendix C for an introduction
to numerical integration using matlab.) We start by putting the equations of motion
in first-order form using the change of coordinates
Z=
z1
y
=
ẏ
z2
,
which implies
ż1 = z2
ż2 = −
c
GM
+ z22.
2
(Re + z1)
m
(2.28)
In matlab, we create the function file falling_ode.m, which contains the first-order
equations of motion from Eq. (2.28):
function zdot = falling_ode(t,z,G,M,m,Re,c)
zdot(1,1) = z(2);
zdot(2,1) = -G*M/(Re+z(1))^2+c/m*z(2)^2;
To integrate these equations, we use the matlab Command Window to define all
relevant input variables and then call ODE45:
>> [t,z] = ode45(’falling_ode’,tspan,z0,[],G,M,m,Re,c);
Figure 2.11 shows a plot of z1(t) = y(t) and z2(t) = ẏ(t) for c = 0.05 kg/m,
y(0) = 1,000 m, ẏ(0) = 0 m/s, m = 10 kg, and t = [0, 20] s. Note that the velocity
ẏ(t) asymptotically approaches a lower limit, known as the terminal velocity.
We can obtain an analytical estimate of the terminal velocity by rewriting the
equation of motion Eq. (2.27) using the simplifying approximation Re y, that is, by
assuming that the earth’s radius is much larger than the height of the mass. In this case,
the inverse square gravity force simplifies to a constant gravitational acceleration,
g = GM
= 9.81 m/s2. Under this assumption, the approximate equation of motion is
R2
e
ÿ ≈ −g +
c 2
ẏ .
m
(2.29)
Note that the approximate equation of motion can also be obtained by assuming that
the force of gravity is equal to the sea-level weight of the mass. Solving for the
(constant) terminal velocity ẏ ∗ using Eq. (2.29) by setting ÿ ∗ = 0 gives
gm
∗
.
ẏ = −
c
(Note we take the negative root, since ẏ < 0.) This estimate is plotted as a dashed
line in the lower panel of Figure 2.11 and shows good agreement with the terminal
velocity obtained numerically.
36
CHAPTER TWO
1,000
y (m)
800
600
400
200
0
5
10
15
20
5
10
t (s)
15
20
y (m/s)
0
–20
–40
–60
0
Figure 2.11 Particle falling under gravity with air resistance. The solid curves are the result
of numerically integrating the equation of motion (Eq. (2.27)). The dashed line in the lower
panel is the analytical estimate of the terminal velocity.
Tutorial 2.6 The Tossed Point Mass
This tutorial introduces two-dimensional motion. In Cartesian coordinates this simply
means finding equations of motion for the x and y directions separately. Let m
represent the mass of a tossed particle with position rP/O = (x, y)I . Rather than
modeling the force that launches the mass, we solve for the in-flight trajectory of the
mass as a function of its initial conditions immediately after the toss. The only force
acting on the mass in mid-air (ignoring drag) is gravity, as shown in Figure 2.10b.
Using Newton’s second law, we obtain two equations of motion (one for each degree
of freedom):
ẍ = 0
(2.30)
ÿ = −g.
(2.31)
Integrating the equations of motion (assuming x(0) = y(0) = 0) gives the following
trajectory for the position of the mass:
x(t) = ẋ(0)t
1
y(t) = ẏ(0)t − gt 2.
2
We can solve for the trajectory y(x) by eliminating time t to obtain
y(x) = ẏ(0)
1
x
− g
ẋ(0) 2
x
ẋ(0)
2
.
(2.32)
Eq. (2.32) shows that the tossed mass takes a parabolic trajectory whose shape is
determined by the initial velocities ẋ(0) and ẏ(0). One such trajectory is illustrated
in Figure 2.12 for ẋ(0) = 1 m/s and ẏ(0) = 10 m/s.
NEWTONIAN MECHANICS
37
6
y (m)
4
2
0
0.0
0.5
1.0
1.5
2.0
2.5
x (m)
Figure 2.12 The tossed point mass. A particle tossed from the origin (0, 0) in the presence
of gravity with initial horizontal speed 1 m/s and initial vertical speed 10 m/s travels in the
parabolic trajectory shown.
In this tutorial, we took the approach of ignoring the details of the force that tosses
the mass, since it acts for a short time only. Instead this force was treated as “setting”
the initial conditions of the equations of motion. This idea is very common—it comes
up frequently in this book.
2.7 Key Ideas
.
Newton’s second law of motion in vector form is
FP =
Id
dt
pP/O = mP IaP/O .
I
Newton’s second law describes the inertial acceleration IaP/O of a point mass
P with mass mP under the influence of the total force FP . The quantity FP is
the vector sum of the forces acting on the particle P . When solving a dynamics
problem, the forces acting on P should be hand-sketched in a free-body diagram.
.
.
.
.
A conserved quantity in dynamics is a scalar or vector function of a particle’s
state that remains constant throughout the trajectory. When such a quantity exists,
it is described by a conservation law.
The number of degrees of freedom of a collection of particles is the total number
of independent coordinates needed to describe the position of every particle. Each
particle has three degrees of freedom minus the number of constraints. That is,
if there are N particles and K constraints, then the total number of degrees of
freedom is M = 3N − K. The total number of degrees of freedom of a system
of particles represents the number of equations of motion needed to describe the
motion of the system.
For every constraint there is a constraint force. These forces are called normal
forces if they are perpendicular to the direction of motion.
All units in this text are in SI; that is, we use the following units and abbreviations:
meters (m), seconds (s), and kilograms (kg). The units of force are newtons (N).
38
CHAPTER TWO
.
.
In this chapter we introduced forces used throughout the book—Coulomb friction, viscous friction, air drag, spring (restoring) forces, and gravity.
The trajectory resulting from the one-dimensional equation of motion
z̈ = −ω02z
is known as simple harmonic motion. This equation arises frequently in
dynamics.
2.8 Notes and Further Reading
The historical references in this and later chapters come from a variety of sources.
A particularly comprehensive discussion of the history of mechanics can be found
in Dugas (1988). For a popular treatment, more information on Galileo and his
contribution to mechanics can be found in Dava Sobel’s wonderful book Galileo’s
Daughter (Sobel 1999). There have been many editions of Newton’s Principia over
the centuries. We used the 1934 University of California edition (Newton 1934). A
particularly good popular book on Newton’s science and the history of the Principia
is Ivar Peterson’s Newton’s Clock, Chaos in the Solar System (Peterson 1993). James
Gleick also has an engaging biography of Newton (Gleick 2003). Our discussion of
the terms “theory” and “law” comes from a publication on evolution and creationism
(National Academy of Sciences 1999).
Many dynamics and physics texts discuss in varying degrees of detail Newton’s
laws and the important concepts of force, mass, acceleration, and absolute space.
For example, introductory descriptions can be found in Meriam and Kraige (2001),
Bedford and Fowler (2002), Tenenbaum (2004), and Thornton and Marion (2004).
A more detailed and particularly clear discussion is found in Greenwood (1988). An
insightful discussion of the concept of force in physics can be found in the 2004
Physics Today column by Frank Wilczek (Wilczek 2004).
Most introductory dynamics texts also discuss units and conversions. For our
discussion, we referred to Meriam and Kraige (2001), Bedford and Fowler (2002),
Hibbeler (2003), Tongue and Sheppard (2005), and Beer et al. (2007). An excellent
reference on standards and units is the website http://www.nist.gov of the National
Institute of Standards and Technology.
2.9 Problems
2.1
Newton’s second law applies to a point mass with the acceleration relative to
what type of frame?
2.2
In Newton’s second law, what does mP refer to?
2.3
Consider the straight-line motion of a particle of mass mP acted on only by
air resistance, FP = −bẋ 2. Find analytically an expression for the velocity of
the particle as a function of time if it starts with initial velocity v0 at time t0.
NEWTONIAN MECHANICS
39
2.4
In Example 2.4 we showed that the number of degrees of freedom for a car
on a planar roller coaster reduces to one. What is the corresponding constraint
force?
2.5
Consider the luge shown in Figure 2.13. Treating the rider and sled as a point
mass, how many degrees of freedom does it have? What coordinates might
you use?
Figure 2.13 Problem 2.5. Image courtesy of Shutterstock.
2.6
Sketch a planar model of a weightlifter’s barbell (shown in Figure 2.14) using
two point masses and a rigid massless rod. How many degrees of freedom are
there in your model? (HINT: Find the constraint equation(s).)
Figure 2.14 Problem 2.6. Image courtesy of Shutterstock.
40
CHAPTER TWO
2.7
In Example 2.5 we found that the three-link robot arm in Figure 2.6a has
three degrees of freedom. We described them by three angles in Figure 2.6b.
Suppose, instead, you desired to describe the system by the six Cartesian
coordinates for the end of each link, (x1, y1)I , (x2, y2)I , and (x3, y3)I . How
many constraint equations would be necessary and what are they?
2.8
Draw the free-body diagram for the barbell model developed in Problem 2.6.
2.9
Draw the free-body diagram for the luge model in Problem 2.5.
2.10 A rocket lifting off from the ground is acted on by a variety of forces, including
thrust, gravity, drag, and wind. Treating the rocket as a point mass, draw the
free-body diagram of the rocket during ascent.
2.11 Despite Galileo’s claim, large, heavy skiers have an unfair advantage: they get
down a mountain faster than lighter skiers. In fact, in the four-person bobsled,
for instance, ballast weights are added to the sled to ensure that every sled,
including the drivers, has the same weight (again giving an advantage to larger
racers, as the lighter competitors have to push a heavier sled at the start).
Where did Galileo go wrong? That is, using Newton’s second law and your
knowledge of the forces at work, show why it is that heavier skiers get down
the mountain faster.
(You may assume that the skier is acted on by gravity; friction, F = μN;
and aerodynamic drag, F = 21 CD Aρv 2.)
2.12 Find the equations of motion for a mass m suspended vertically from a spring
as shown in Figure 2.15, assuming that the mass is constrained to move only
vertically and that it is subject to the force of gravity. Draw a free-body
diagram, choose a coordinate system, and use Newton’s second law to find
the equation of motion.
g
m
Figure 2.15 Problem 2.12.
2.13 Many springs are what are called nonlinear springs. That is, the force depends
nonlinearly on position. A common nonlinear-spring force is F = −k(x −
x0) − c(x − x0)3, where x0 is the unstretched length of the spring, and k and
c are constants. One way to think about F is that the spring gets stiffer as it is
stretched.
a. Find the equation of motion for a mass on a frictionless surface attached to a linear spring (c = 0) and then for a nonlinear spring, and
put each equation in first-order form.
b. Numerically integrate the equations of motion over the time interval
[0, 20] s. Assume the unstretched length of the spring x0 is 0.25 m.
NEWTONIAN MECHANICS
41
Assume an initial stretched position of 0.4 m. Use k = 1 N/m, c = 5
N/m3, and m = 1 kg. Plot the resulting trajectories on the same axes
and compare the responses.
2.14 Show that the general solution for a critically damped harmonic oscillator is
given by Eq. (2.26). Find expressions for the constants A and B in terms of
the initial conditions z(0) and ż(0).
2.15 Find expressions for the position x(t) as a function of time for a critically
damped spring and an overdamped one in terms of the initial conditions x(0)
and ẋ(0).
2.16 A rock dropped from the top of a cliff will experience an acceleration due to
gravity along with a deceleration due to drag. The downward acceleration is
g − cẏ 2, and the downward speed is ẏ. If the rock is dropped from a height of
y m, calculate the rock’s impact velocity. How high will the cliff have to be
for the rock to hit the ground at 99% of its terminal velocity? (This problem
should be solved analytically, that is, without the use of a computer.)
2.17 Consider the equation of motion of a driven simple harmonic oscillator
ẍ = a −
k
x,
m
where a is a constant.
a. Integrate the equation to find the solution analytically.
b. Solve the equation of motion using matlab ODE45, and plot x(t) and
ẋ(t). Let a = 1, m = 0.5, and k = 3. Be sure to label your axes (with
units!) and add a legend to the plot.
2.18 In Tutorial 2.5 we solved for the equation of motion describing a particle
falling under Newton’s law of gravity and acted on by air resistance. We found
that the particle reached terminal velocity. One approximation made, however,
was that the density of the earth’s atmosphere is constant over the trajectory.
A better model of atmospheric density includes exponential decay,
ρ = ρ0e−y/ h,
where ρ0 is the atmospheric density at sea level (roughly 1.2 kg/m3), and h is
called the scale height of the atmosphere. Near the surface, h is approximately
7,000 m (the actual value can vary with temperature and season). Using the
same parameter values as in Tutorial 2.5 and assuming the given value of c
at sea level, find the new equation of motion incorporating the exponential
atmosphere model and simulate the resulting height and speed trajectories.
Compare the terminal velocity you find to that in Figure 2.11.
2.19 Two masses are released from a height h above the ground as shown in Figure 2.16. Mass P is released from rest and mass Q is launched horizontally
with speed v0. Which mass lands first? Draw a free-body diagram for each
42
CHAPTER TWO
P Q v0
g
h
Figure 2.16 Problem 2.19.
mass, choose a coordinate system, use Newton’s second law to find the equations of motion for each mass, and solve the equations of motion for the time
until impact with the ground.
PART ONE
Particle Dynamics in the Plane
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CHAPTER THREE
Planar Kinematics and Kinetics of a Particle
Chapter 2 introduced the fundamental physical concepts and tools for solving all
dynamics problems. In it, we also solved complete and important problems in one
dimension. To whet your appetite, toward the end of the chapter we had a first look
at generalizing Newton’s second law to two dimensions, though using only a scalar
notation (as opposed to the vector notation developed here), by solving for the motion
of a point mass under gravity. Although it is true that all particle dynamics problems
could be solved using only Cartesian coordinates and simple extensions of the scalar
treatment used in the last chapter, that would be unnecessarily limiting. This chapter
introduces the vector formalism used for modern treatments of dynamics and lays
out the basic mathematical foundation we use throughout the book. By the end you
will have almost all the tools you need to solve any dynamics problem. We limit
the discussion to only two dimensions, however, to focus on the important concepts
without overly complicating the mathematics. Part Four shows that only a few new
ideas are necessary to extend this analysis to three dimensions.
3.1 The Simple Pendulum
The limitations of what we have learned up to now are easily demonstrated by an only
slightly more complicated example—the simple pendulum, illustrated in Figure 3.1.1
In this example, we provide a first look at the problem-solving notation and techniques
that are used throughout the rest of the book. Don’t worry if these tools seem a bit
mysterious at first; things should clarify by the end of the chapter.
1 The pendulum is simply a swinging mass in a uniform gravity field. Its periodic motion was first discovered
by Galileo Galilei in 1602. Pendulums have been incorporated into many devices since, most notably clocks
and accelerometers. The simple pendulum refers to a pendulum consisting of a point mass on a massless
rod or string, whereas the compound pendulum refers to a pendulum consisting of an extended rigid body.
Figure 3.1 is a simple pendulum in a configuration that is sometimes referred to as an inverted pendulum
because it is directed up rather than down.
46
CHAPTER THREE
P
I
F
P
g
Fy
–
l
P
l
y
O
Fx
–
l
O
W
x
(a) Coordinates
(b) Reference frame
(c) Free-body diagram
Figure 3.1 Solving the pendulum using Cartesian coordinates in an inertial frame.
Example 3.1 Solving the Pendulum Using Cartesian Coordinates
in the Inertial Frame
Consider a mass mP fixed to a stiff (massless) rod of length l that pivots freely (in
the plane of the page) about the point O, as shown in Figure 3.1. We approach this
problem in the same manner as the tossed point mass in Tutorial 2.6. The Cartesian
coordinates of the pendulum mass mP with respect to the inertial frame I are (x, y)I .
Just as in Tutorial 2.6, we write two equations of motion, one in the x-direction and
one in the y-direction, in terms of the forces acting in those directions and the speeds
and accelerations, (ẋ, ẏ)I and (ẍ, ÿ)I , respectively. Using the free-body diagram
in Figure 3.1c, Newton’s second law applied to each of the two planar directions
is then
Fx
= mP ẍ
l
Fy
− mP g = mP ÿ.
l
(3.1)
(3.2)
The force W in Figure 3.1c is the weight of the mass, which acts downward
with magnitude mP g. The force F is the internal tension or compression in the
rod connecting the mass to the pivot. (Recall that by Newton’s third law, the rod
experiences an equal and opposite force.) Because the rod is massless and can pivot
freely about point O, there is no rod-mass interaction force that is perpendicular to the
rod—the only force from the rod on the mass is along the line between point O and
the mass.2 (In general, F can be positive or negative.) To write Eqs. (3.1) and (3.2), we
have used our knowledge of vector sums and free-body diagrams to write the tension
as the sum of two other forces of magnitudes F x/ l and F y/ l, acting in the x- and
y-directions, respectively. Nevertheless, there is only one unknown magnitude, F .
To derive the equations of motion, we eliminate F from Eqs. (3.1) and (3.2) to
obtain the single equation
ẍy − ÿx = gx.
(3.3)
2 This observation does not hold if we were to model the pendulum rod as having finite mass (i.e., as a
rigid body).
PLANAR KINEMATICS AND KINETICS OF A PARTICLE
47
The pendulum is an example of a single-degree-of-freedom system (see Section 2.4)
and thus has a constraint on the position,
x 2 + y 2 = l 2.
(3.4)
Differentiating the constraint in Eq. (3.4) with respect to time and using the fact that
l is constant yields
x ẋ + y ẏ = 0.
(3.5)
Differentiating Eq. (3.5) a second time yields
ẋ 2 + ẏ 2 + x ẍ + y ÿ = 0.
(3.6)
Solving Eqs. (3.3) and (3.6) for the two unknowns ẍ and ÿ, we obtain
ẍ = (−x ẋ 2 − x ẏ 2 + xyg)/ l 2
ÿ = (−y ẋ 2 − y ẏ 2 − x 2g)/ l 2.
(3.7)
Note that we explicitly used the constraint Eq. (3.6) to arrive at Eq. (3.7), so the
solution automatically satisfies Eq. (3.4).
Solving Eq. (3.7) will certainly give the correct motion for the pendulum in
Cartesian coordinates, but not only are these equations complicated (there is no exact
solution), they may also be very poorly behaved numerically. You have a chance
to explore their numerical behavior in Problem 3.22 at the end of the chapter. The
equations of motion also provide no obvious qualitative insight into the behavior of
the pendulum. For instance, your experience should tell you that the way we drew
the pendulum in Figure 3.1 results in unstable motion; that is, the pendulum will fall
until hanging downward. We call this configuration an inverted pendulum.
All is not lost, however. With a few new concepts and some careful notation, we can
dramatically simplify the approach to solving this problem and many others. These
new concepts and notation are the subject of the remainder of this chapter.
We start by introducing some important concepts from vector algebra and vector
calculus. Pay particular attention to the notation—it will be used throughout the book,
and being comfortable with it is essential. Although some of these concepts may be
familiar to you, we still carefully define each one to develop a consistent approach.
3.2 More on Vectors and Reference Frames
3.2.1 The Vector
In Chapter 1 we gave a qualitative definition of the term “vector.” Understanding this
concept and how we use it in mechanics is essential. The vector is the fundamental
geometric quantity used to tie the physical world to the mathematics of dynamics.
The vector is the foundation for all that we do in solving mechanics problems—it
thus helps to be familiar with the algebraic properties of vectors. For example, the
concepts of total force and the free-body diagram discussed in Chapter 2 use the
48
CHAPTER THREE
additive properties of vectors. In case you are rusty on the properties and rules for
vectors, Appendix B summarizes some useful identities.
We make frequent use in the book of the unit vector. The unit vector is a vector
that is constrained to have unit length. That is, a unit vector has length equal to one.
Introducing the unit vector lets us write an arbitrary vector as a scalar magnitude
times a unit vector pointing in the same direction, which can be convenient for many
problems. We label all unit vectors with lowercase, boldface, roman type, and we
often use the letter e. Unit vectors often have a subscript arabic numeral or letter
(e.g., e1 or ex ). A carat above a vector identifies a unit vector pointing in the same
direction. For example, given a vector r, the unit vector in the same direction is
r̂ =
r
,
r
where the notation r denotes the magnitude of the vector r.
We also often label unit vectors by the lowercase letter corresponding to the
uppercase label of the reference frame in which they are fixed, such as the unit vectors
b1, b2, b3, which are fixed in frame B. In fact, the most important use of a unit vector
is to define the directions in a reference frame, which we discuss next.
3.2.2 The Reference Frame
Chapter 1 presented a qualitative definition of the reference frame. That qualitative
definition allowed us to introduce what we mean by dynamics and to discuss the
important concept of absolute space in the context of Newton’s laws. The qualitative
definition also made it possible to give abstract meaning to the rate of change of a
vector. Here we present two equivalent definitions of a reference frame—a physical
definition and an operational one. These new definitions of a reference frame permit
solving a broader class of problems by tying physical concepts to the mathematical
tools used.
Physical Definition 3.1 A reference frame is a collection of at least three noncollinear points that are rigidly connected or rigidly attached; that is, the distance between any two points in the collection does not change with time.
The physical definition of a reference frame points out that frames are attached
to entities and can thus move and rotate with respect to one another. For example,
an inertial frame is fixed with respect to absolute space. A rigid body, which is a
collection of particles with fixed relative spacing from one another, can be used to
define a reference frame. When we study rigid bodies in Parts Three and Four, the
interchangeability of a rigid body and the reference frame associated with it will be
indispensable. Like rigid bodies, reference frames cannot change shape. They don’t
distort, expand, or contract. The orthogonal unit vectors defining the frame stay fixed
with respect to one another, which leads us to the third equivalent definition of a
reference frame.
Operational Definition 3.2 A reference frame is defined by three orthogonal
unit vectors and one point (the origin).
In this text, a reference frame is labeled by a calligraphic capital letter and defined
according to Definition 3.2. For example, we write I = (O, e1, e2, e3) to define
PLANAR KINEMATICS AND KINETICS OF A PARTICLE
I
49
e2
O
e1
e3
Figure 3.2 Reference frame I = (O, e1, e2 , e3) is defined by its origin O and three orthogonal
unit vectors e1, e2 , and e3.
reference frame I using the origin O and the unit vectors e1, e2, and e3. Frame I
is illustrated in Figure 3.2, where e3 = e1 × e2 (see Appendix B).
The operational definition of a reference frame provides the elements needed to
perform operations on vectors. As discussed in Chapter 1, the concept of a reference
frame is essential to dynamics. How we describe a vector—and how its motion is
defined—differs depending on the frame in which we are interested. Most importantly,
Newton’s laws of dynamics imply a specific reference frame. Recall from Chapter 2
that Newtonian dynamics applies only when the acceleration is referred to a special
reference frame called the inertial frame. When using Newton’s second law to solve
for the motion of mass P , we use the acceleration IaP/O —the second time derivative of
the position rP/O with respect to the inertial frame. The vector quantity IaP/O indicates
how the velocity IvP/O evolves relative to the inertial frame. Newton’s second law can
be written and solved in coordinates defined in any frame, but the vector acceleration
must refer to the time rate of change with respect to an inertial frame. We return to
this important point later.
Parts One through Three consider dynamics problems in which the motion is
confined to a plane spanned by two of the three unit vectors. Nonetheless, it is
important to keep track of the third unit vector, as it plays a key role in solving
problems using Newton’s method.3 We treat three-dimensional motion in Part Four.
Once you deeply understand the idea of the reference frame, you can move on to
the concept of vector components.
3.2.3 Components of a Vector
We often write a vector as the weighted sum of the unit vectors of a reference frame.
For instance, if reference frame A is described by the three unit vectors a1, a2, and
a3, then the position of point P relative to O can be written
rP/O = a1a1 + a2a2 + a3a3.
3 If the third unit vector is not provided in the frame definition, we assume that it equals the cross product
of the first unit vector with the second unit vector.
50
CHAPTER THREE
P
B
b2
A
b2b2
rP/Q
a2
b1b1 a2a2
Q
a1a1
b1
O
a1
a3, b3
Figure 3.3 Vector components of rP/Q in reference frames A = (O, a1, a2 , a3) and B =
(O, b1, b2 , b3).
These weighted unit vectors—a1a1, a2a2, and a3a3—are the components of the vector
rP/O in frame A. The scalar ai , i = 1, 2, 3, is the magnitude of the component of rP/O
in the direction ai . We can also write rP/O in terms of components in a second reference
frame B:
rP/O = b1b1 + b2b2 + b3b3.
Figure 3.3 shows this idea graphically for the case a3 = b3.
We can also write the magnitudes of each component of a vector in matrix notation.
For example, we can describe the vector rP/O in frame A as
⎡
⎤
a1
[rP/O ]A = ⎣ a2 ⎦ .
a3 A
Note that we indicate the reference frame to which the scalars refer. Matrix notation
eliminates the need to explicitly write the unit vectors—these vectors are implied
by the frame definition. Matrix notation also allows replacement of some vector
operations with matrix operations. Matrix operations become particularly useful when
discussing three-dimensional motion of rigid bodies in Part Four.
The ability to write a vector as a weighted sum of the unit vectors of a specific
reference frame is extremely useful. You should become very comfortable with this
notation.
Example 3.2 The Pendulum Position Expressed in Vector Components
We are now in a position to treat the pendulum a bit more carefully using vectors.
In this example the position of the pendulum is represented as a vector, rather than
just using the scalar Cartesian coordinates as in Example 3.1. Figure 3.4 shows the
simple pendulum and two frames of reference with origin O: the inertial frame
I = (O, ex , ey , ez ) and a second frame B = (O, b1, b2, b3), oriented so the b1 axis
PLANAR KINEMATICS AND KINETICS OF A PARTICLE
I
51
b1
ey
P
b2
B
ex
O
Figure 3.4 The pendulum represented by components in two different frames.
is always pointed at the pendulum bob. The position of the pendulum bob P can be
written as components in the inertial frame I :
rP/O = xex + yey ,
where as before, (x, y)I must satisfy the constraint that x 2 + y 2 = l 2.
However, when written as components in frame B, the position is
rP/O = lb1.
As shown later, writing the pendulum as components in this frame can significantly
simplify the problem of finding the equations of motion.
Example 3.3 Satellite Tracking
Figure 3.5 shows an inertial frame I located at the center of the earth. A second
frame, B, is located at a position on the surface of the earth at 45◦ latitude, such
that b1 and e1 are aligned, b3 is aligned with the vertical at O , and b2 is along the
horizon to complete an orthogonal right-handed set (b2 = b3 × b1). Suppose we are
tracking a satellite from a station located at O , and we wish to know the position of
the satellite, S, relative to the center of the earth, O, expressed as components in the
inertial frame I .
I
b3
e3
B
O′
R
O
b2
45°
e2
e1
rS/O′ = hb2
S
Figure 3.5 Satellite tracking from the earth’s surface.
52
CHAPTER THREE
As shown in Figure 3.5, at the time of measurement the satellite is just on the
horizon, so its position in terms of components in frame B is
rS/O = hb2,
where h is the range to the satellite.
The position of the satellite relative to O is given by the addition rule for vectors,
rS/O = rO /O + rS/O = Rb3 + hb2,
where R is the radius of the earth.
To find the satellite position entirely as components in frame I , we need to write
the vectors b2 and b3 as components in I . This is done by inspection:
1
1
b2 = √ e2 − √ e3 and
2
2
1
1
b3 = √ e2 + √ e3.
2
2
(3.8)
The final position of the satellite as components in I is thus
1
1
rS/O = √ (R + h) e2 + √ (R − h) e3.
2
2
3.2.4 Coordinate Systems
The next concept of great importance is the coordinate system. Recall that we defined
a coordinate system in Chapter 1 as a set of scalars that locates the position of a point.
We strongly emphasized the distinction between a coordinate system and a reference
frame—they are not the same thing! A coordinate system locates a point in a specific
reference frame. This section introduces three essential coordinate systems that are
used to describe the configuration of points in the plane. We introduce two more
coordinate systems when discussing three-dimensional motion in Part Four.
Cartesian Coordinates
The most common coordinate system is the one we have already been using—
Cartesian coordinates. The Cartesian coordinates of an arbitrary frame I are equal
to the magnitudes of the components of a vector expressed in frame I . The letters
x, y, and z are often used to label Cartesian coordinates. For example, the position
of point P relative to O expressed using Cartesian coordinates in reference frame
I = (O, ex , ey , ez ) is
rP/O = xex + yey + zez .
PLANAR KINEMATICS AND KINETICS OF A PARTICLE
I
53
e2
P
r
θ
O
e1
Figure 3.6 Polar coordinates (r, θ )I locate the position of point P relative to point O.
In matrix notation, we have
⎡
⎤
x
[rP/O ]I = ⎣ y ⎦ .
z I
Cartesian coordinates are just one possible choice of coordinates. For some problems, particularly problems with constraints, another coordinate system may simplify
the equations of motion. A judicious choice of coordinate system (and corresponding
reference frame) may allow the use of fewer scalars to represent the motion. Part of
the art of solving dynamics problems is selecting the coordinate systems and reference
frames that maximize insight and minimize algebra.
Polar Coordinates
One alternative to the Cartesian coordinate system is the polar coordinate system.
In polar coordinates a point P is located in a plane by the polar radius r and the
polar angle θ, as shown in Figure 3.6. We can still write the vector rP/O in terms of
components in the e1 and e2 directions in reference frame I , only we use the scalars
associated with polar coordinates:
rP/O = r cos θe1 + r sin θe2.
(3.9)
In matrix notation, we have
r cos θ
[rP/O ]I =
r sin θ
.
I
Path Coordinates
Another type of coordinate system is the path coordinate system. Path coordinates
are also called curvilinear coordinates or normal/tangential coordinates. Unlike
the Cartesian and polar coordinate systems, which use two scalars to locate a
point P in a plane, the path coordinate system uses only a single scalar s to locate
P along a curve. Thus path coordinates are typically used for problems that have only
one degree of freedom. Clearly, the value of s is not enough information to completely
express the position of P relative to O. Some mathematical description of the path is
54
CHAPTER THREE
I
P
e2
s
O
e1
Figure 3.7 An arbitrary path in the fixed reference frame I and path coordinate s measured
along the path.
also necessary (e.g., the oval curve that describes a racetrack). Figure 3.7 sketches a
situation in which path coordinates may be the best choice.
Example 3.4 The Simple Pendulum Using Different Coordinates
Example 3.2 described the position of a pendulum bob in Cartesian coordinates in an
inertial reference frame I = (O, ex , ey , ez ) (and found the corresponding equations
of motion). Here we look again at the pendulum and describe its position using polar
and path coordinates.
Figure 3.8 shows a pendulum and an inertial reference frame I = (O, e1, e2, e3).
Note that we have changed the definition of the inertial frame from that in Example 3.2: the directions of the axes are arbitrary as long as the frame remains inertial.
(We often move the inertial frame around to find the most convenient coordinate system for a problem.) This configuration is more typical for the simple pendulum in
contrast to the inverted pendulum shown in Figure 3.1. The pendulum-mass position
in Cartesian coordinates is still given by (x, y)I , where x is the coordinate in the e1
direction and y is the coordinate in the e2 direction, and the length constraint is still
x 2 + y 2 = l 2.
A Cartesian coordinate system is not the most convenient for the simple pendulum,
however, because it uses two scalar coordinates for the single-degree-of-freedom
system; we should be able to find a single coordinate to describe the pendulum’s
configuration. Figure 3.8 uses polar coordinates to describe the configuration. With
polar coordinates, we can completely describe the position of the pendulum bob by
e2
O
r
θ
l
e1
I
s
P
Figure 3.8 The simple pendulum described by polar and path coordinates in reference frame
I . The position of pendulum bob P is entirely determined by polar angle θ or path length s.
PLANAR KINEMATICS AND KINETICS OF A PARTICLE
55
the polar angle θ. The polar coordinate r is equal to the (fixed) length of the pendulum
rod. (We solve the pendulum problem using polar coordinates in Example 3.9.) The
polar coordinate θ is related to the Cartesian coordinates by
y
.
θ = arctan
x
We can also describe the position of the pendulum as the distance along the arc
of a circle of radius l, as indicated by the single path coordinate s in Figure 3.8. The
path length s is related to the polar angle θ by
s = lθ.
In each case, we are still describing the position of the pendulum mass in reference
frame I . These are three equivalent ways of doing so using the three coordinate
systems introduced so far.
Example 3.5 An Inclined Plane
Figure 3.9 shows a box of mass mP sliding down an inclined plane of height h and
base b. An inertial frame I is aligned with the base and has its origin at the lower left
corner. What is the position of the box in the three coordinate systems studied so far?
The box position in Cartesian coordinates in I is (x, y)I . However, this problem
has one degree of freedom, since the box is able to slide only along the incline. This
implies the following constraint:
y
1
= tan 30◦ = √ .
b−x
3
We could also specify the position of the box in polar coordinates, also shown in
Figure 3.9. The polar coordinates are related to the Cartesian coordinates by
x2 + y2
y
θ = arctan
x
r=
,
I
e2
s
P
h
r
y
θ
O
x
30°
e1
b
Figure 3.9 Block on an inclined plane in three coordinate systems.
56
CHAPTER THREE
The constraint in polar coordinates becomes
b
.
r=√
3 sin θ + cos θ
The single degree of freedom also implies that we can locate the box using only
one scalar coordinate. In this case, the most convenient coordinate is a path coordinate
s, specifying the distance the block has slid down the incline, as shown in Figure 3.9.
The relationship of the Cartesian coordinates to s is
√
◦
x = s cos 30 = s
3
2
s
y = h − s sin 30◦ = h − ,
2
√
where l is the length of the incline, l = h2 + b2. We could likewise find similar
relationships between the polar coordinates and the path coordinate (see Problem 3.4).
It is worth noting that the reason the relationships between s and the other coordinates (and the constraints) are complicated is due to our choice of reference frame.
We could have instead chosen a frame located at the top of the incline and rotated
so that the e1 direction is along the incline. In that case the path coordinate would
be the same as the Cartesian x coordinate and the polar coordinate r. The constraints
would be trivial: y = 0 for Cartesian coordinates and θ = 0 for polar coordinates. This
example highlights the importance of a careful choice of reference frames.
3.3 Velocity and Acceleration in the Inertial Frame
Now that you are comfortable with vectors, reference frames, and coordinate systems,
we are ready to discuss the vector derivative. The vector derivative is fundamental to
the topic of kinematics, which describes the motion of particles (and rigid bodies)
without reference to the forces acting on them. In this section we carefully define the
vector derivative and derive certain properties. Then we use the vector derivative to
find the velocity and acceleration of a point in the inertial frame in terms of the three
coordinate systems studied so far.
3.3.1 The Vector Derivative
Chapter 1 qualitatively described the vector derivative as the change with respect to
time of the magnitude and direction of a vector in a reference frame. Figure 3.10 again
illustrates this qualitative description, now using our notations for the unit vectors and
origin of a reference frame. Remember that the time derivative of a vector is itself a
vector. In the sections that follow, we provide a mathematical definition of the vector
derivative, state its properties (we prove them in Section 3.8), and show its equivalence
to our geometric ideas. If you are rusty, you can review the definition of the scalar
derivative of a function in Appendix A.
PLANAR KINEMATICS AND KINETICS OF A PARTICLE
I
e2
P
57
I
d (r ) = Iv
—
P/O
P/O
dt
rP/O(t)
rP/O(t + Δt)
O
e3
e1
Figure 3.10 Schematic representation of the vector derivative.
Definition 3.3 Let r(t) = r1(t)a1 + r2(t)a2 + r3(t)a3 be a vector function of time
expressed as components in reference frame A = (O, a1, a2, a3). If the scalar
functions r1(t), r2(t), and r3(t) are all differentiable at t = a, then the vector
function r(t) is differentiable in reference frame A at t = a with the vector
derivative in A given by
Ad
dt
=
r
t=a
d
d
r1
a1 +
r2
dt t=a
dt
a2 +
t=a
d
r3
dt
a3
(3.10)
t=a
or, more compactly,
Ad
dt
= ṙ1(a)a1 + ṙ2(a)a2 + ṙ3(a)a3.
r
t=a
As emphasized in Chapter 1, it is essential to remember that the derivative of a
vector may be different in different frames. The change in a vector—magnitude or
direction—has no meaning unless a reference frame is specified, since the frame may
also be moving. (The change in a scalar, however, does not need a frame specification.)
This stipulation should hopefully be clear from our definition. Since the unit vectors
a1, a2, a3 define frame A, they are fixed in it and thus the change with time of r(t) in
A is given by Eq. (3.10). However, if we asked how r changed in a different frame,
say B, then the ai unit vector might be changing because B might be rotating with
respect to A.
As an example, consider a point P traveling around a circle centered at O, as in
Figure 3.11. The rate of change of rP/O with respect to a fixed reference frame I
with origin O is simply the change in direction of rP/O as P travels around O. The
corresponding velocity is written IvP/O and is always tangent to the circle. However,
the rate of change of rP/O with respect to a rotating frame B = (O, b1, b2, b3), where
b1 is always aligned with rP/O , is zero. Point P is not moving in B!
This example highlights the importance of labeling a vector derivative with the
appropriate reference frame. We always specify the frame to which a velocity is related
I
by using a superscript capital letter for that frame (e.g., IvP/O = dtd rP/O ). Likewise
58
CHAPTER THREE
I
B
I
e2
vP/O
b2
P
rP/O
b1
O
e1
Figure 3.11 Point P traveling around a fixed circle with velocity IvP/O tangent to the circle.
In a frame B rotating so that b1 always points at P , the velocity is zero: BvP/O = 0.
Table 3.1
Properties of the vector derivative
Addition rule
Product rule
Ad
dt
Ad
dt
(r(t) + s(t)) =
(f (t)r(t)) =
df
dt
Ad
dt
r(t) +
Ad
dt
s(t)
A
(t)r(t) + f (t) dtd r(t)
Ad
d
.
.
. Ad
dt (r(t) s(t)) = dt r(t) s(t) + r(t) dt s(t)
Ad
dt
Chain rule
Ad
dt
Ad
dt
(r(t) × s(t)) =
r(f (t)) =
Ad
dt
r(t) × s(t) + r(t) ×
Ad
dt
s(t)
Ad
df
df r(f (t)) dt
f (r(t)) = ∇f .
Ad
dt
r
I
for an acceleration, we write IaP/O = dtd (IvP/O ). Although this notation may seem
burdensome, writing vector derivatives this way is an important habit to begin now.
Remember that Newton’s second law only applies when the acceleration (and the
velocity it is differentiating) is defined with respect to an inertial frame.
It turns out that all important properties of the scalar derivative also apply to vector
derivatives; they are summarized in Table 3.1. (If you are rusty on the scalar versions,
you can review them in Appendix A.) These properties are often used in our later work,
particularly when finding expressions for the inertial velocity and inertial acceleration
of a particle in terms of the three coordinate systems studied so far. These rules are
derived in Section 3.8.
Example 3.6 The Vector Derivative on an Inclined Plane
This example returns to the block sliding on the inclined plane of Example 3.5, as
shown again in Figure 3.12. Here we have labeled the block by the letter P and
included three additional frames, an inertial frame A aligned along the plane and
fixed to the upper corner, a frame B fixed at O whose b1 direction always points at P ,
and a frame C fixed to the block and moving with it with direction c1 along the incline.
PLANAR KINEMATICS AND KINETICS OF A PARTICLE
I
A
a2
e2
C
c2
O′
s
P
B
b2
59
b1
h
r
y
θ
O
30°
e1
b
x
a1,c1
Figure 3.12 Block on an inclined plane and four different reference frames.
The origin of C is O = P . From Example 3.5, the position of the block expressed as
components in each frame and in terms of the different coordinates is
rP/O = xe1 + ye2 = r cos θe1 + r sin θe2
= rb1
and
rP/O = sa1.
We now use the definition of the vector derivative to find the velocity of P in each
frame. From the geometry it is obvious that the velocity of P in C is zero, as it always
remains at the origin of C:
Cd
dt
rP/O = 0.
The velocity of P in A is given by the vector derivative
Ad
dt
rP/O = ṡa1.
The velocity of P in I is given in terms of the Cartesian coordinates as
Id
dt
rP/O = ẋe1 + ẏe2
and in terms of polar coordinates as
Id
dt
rP/O = (ṙ cos θ − r θ̇ sin θ)e1 + (ṙ sin θ + r θ̇ cos θ)e2,
60
CHAPTER THREE
where we have used the chain rule for vector derivatives. Note that we can use the
constraints described in Example 3.5 to rewrite the vector derivative in I in terms of
the single coordinate s describing the single degree of freedom:
√
Id
1
3
rP/O = ṡ
e1 − ṡ e2.
dt
2
2
Finally, the velocity of P in frame B is again given by the definition of the vector
derivative:
Bd
dt
rP/O = ṙb1.
3.3.2 Velocity Is the Vector Derivative of Position
Now that we have a formal mathematical definition of the vector derivative, it is useful
to show how the qualitative description follows from our formal definition. We use
the definition of the scalar derivative in Definition A.2 to rewrite the vector derivative
in Eq. (3.10) as
Ad
dt
=
r
t=a
lim
h→0
r1(a + h) − r1(a)
h
a1 +
lim
h→0
+
lim
h→0
r2(a + h) − r2(a)
h
a2
r3(a + h) − r3(a)
h
a3.
(3.11)
We can now use the distributive property of the limit operator and of vector addition
to obtain
Ad
dt
= lim
r
t=a
h→0
1
r1(a + h)a1 + r2(a + h)a2 + r3(a + h)a3
h
− r1(a)a1 + r2(a)a2 + r3(a)a3 .
Thus the vector derivative is simply the limit of the difference between the vector
evaluated at time t = a in frame A and the vector a short time t = a + h later—
also expressed in frame A—just as illustrated in Figure 3.10. Again, it is important
to remember that we evaluate the change in the vector by expressing it in terms of
components in frame A at both times. The vector derivative—or any change in a
vector—is frame dependent.
The vector derivative of the position of point P with respect to an inertial frame I is
the inertial velocity of P , illustrated in Figure 3.10. Eq. (3.11) shows that the velocity
always points along the tangent to the trajectory of P , as illustrated in Figure 3.11. The
trajectory of P —traced by the tip of the vector rP/O —is defined by the three scalar
functions r1(t), r2(t), and r3(t). From Appendix A and Eq. (3.11), the magnitudes
of the components of the velocity IvP/O are the slopes of those three functions. The
velocity is thus directed along the line defined by these three slopes, that is, the tangent
to the trajectory.
We could go through the same development by replacing rP/O (t) with IvP/O in
Definition 3.3 to derive the acceleration IaP/O and show that it is tangent to the curve
PLANAR KINEMATICS AND KINETICS OF A PARTICLE
61
traced by the tip of the velocity. It is not necessary, however, because our development
of the vector derivative was completely general—it applies to any vector, including
the velocity. This is important, since we use the acceleration in Newton’s second law
to obtain the equations of motion needed to solve for the trajectory of P .
3.3.3 Velocity and Acceleration in Various Coordinates
The previous two subsections supplied all the tools you need for vector calculus—in
particular, the vector derivative and its properties. We also showed geometrically how
it is used to define the velocity and acceleration. As discussed in Chapter 1, once the
position of a particle (as a function of time) and its velocity have been determined, we
know everything about its state at all times and can draw its trajectory. The acceleration
allows us to solve for that trajectory using the physics of Newton’s second law.
In reality, although the geometric description is exceptionally important, it is
not as useful as it might at first seem. This is because to actually write down or
draw the position and velocity of a particle we need coordinates. Thus deriving
Newton’s second law for a particle is really about finding differential equations for
the coordinates that describe the position of a particle in a reference frame. In this
subsection we complete the picture by using the definition of the vector derivative
to find expressions for the inertial velocity and acceleration of a particle in the three
coordinate systems discussed so far—Cartesian, polar, and path coordinates.
Kinematics in Cartesian Coordinates
As discussed in Section 3.2.4, Cartesian coordinates are the most common and
fundamental of our coordinate systems. Cartesian coordinates of a point in a reference
frame directly correspond to the magnitudes of the components of the position in that
frame. Thus it is trivial to write the velocity and acceleration of the particle in Cartesian
coordinates, since they follow directly from the definition of the vector derivative:
I
vP/O =
I
aP/O =
Id
dt
Id
dt
rP/O = ẋex + ẏey + żez
vP/O = ẍex + ÿey + z̈ez .
I
Many problems can be solved using Cartesian coordinates. The examples in Chapter 2 all used such coordinates in an inertial frame. Here we have introduced a useful
formalism for thinking about these problems using vectors. While it may seem that
this is superfluous, hang on. To solve more complicated and difficult problems, we
rely on the careful definitions and notations of this chapter.
Example 3.7 The Tossed Point Mass, Revisited
Tutorial 2.6 introduced our first two-dimensional dynamics problem, the tossed point
mass. However, we cheated a bit and avoided the use of vectors, as the important
definitions and notations of this chapter had not yet been introduced. We simply
solved it by means of two separate, scalar one-dimensional systems. Here we revisit it
62
CHAPTER THREE
I
P
e2
rP/O
O
y
e1
x
(a) Coordinates and reference frame
–mPge2
(b) Free-body diagram
Figure 3.13 Solving for the vector motion of a tossed point mass.
and solve it using vector notation. By tackling such an admittedly straightforward
problem, we hope to get you comfortable with the general vector approach to Newton’s second law.
Figure 3.13a shows a picture of the geometry for the example. The position
of the point mass relative to origin O of inertial frame I is given by Cartesian
coordinates, so that rP/O = xe1 + ye2. The point mass P is thrown with an initial
velocity IvP/O (0) = ẋ(0)e1 + ẏ(0)e2. The objective is to solve for the trajectory of P
as a function of time in Cartesian coordinates, rP/O (t) = x(t)e1 + y(t)e2.
The acceleration of the mass in Cartesian coordinates is
I
aP/O =
Id
dt
(ẋe1 + ẏe2) = ẍe1 + ÿe2.
(3.12)
The only force acting on the point mass, shown in the free-body diagram in Figure 3.13b, is the downward action of gravity,
FP = −mP ge2.
Using Newton’s second law, then, we have the vector equation of motion:
I
aP/O = −ge2.
(3.13)
Combining Eq. (3.12) with Eq. (3.13) gives the two scalar differential equations for
the Cartesian coordinates of the point mass:
e1 :
ẍ = 0
e2 :
ÿ = −g.
These are the same two scalar equations as Eqs. (2.30) and (2.31). The rest of the
example is solved as in Tutorial 2.6.
Example 3.8 Solving the Pendulum Using Cartesian Coordinates
in the Inertial Frame, Revisited
We now return to Example 3.1 and use our new notational tools on the simple
pendulum. As shown in Example 3.2, the position of the pendulum bob mP with
PLANAR KINEMATICS AND KINETICS OF A PARTICLE
63
respect to O can be written in terms of components in frame I as
rP/O = xe1 + ye2.
(3.14)
Taking the time derivative with respect to the inertial frame of the position in Eq. (3.14)
by using Definition 3.3 gives the velocity of P with respect to the inertial frame:
I
vP/O =
Id
dt
(xe1 + ye2) = ẋe1 + ẏe2.
(3.15)
Similarly, taking the time derivative with respect to the inertial frame of the velocity
in Eq. (3.15) gives the acceleration of P with respect to the inertial frame:
I
aP/O =
Id
dt
(ẋe1 + ẏe2) = ẍe1 + ÿe2.
(3.16)
Next, using the inertial acceleration in Eq. (3.16) and the free-body diagram in
Figure 3.1c, we can write down Newton’s second law for the pendulum bob:
Fy
Fx
e2 − mP ge2 = mP (ẍe1 + ÿe2).
e1 +
l
l
(3.17)
Vector equation Eq. (3.17) is equivalent to the two scalar Eqs. (3.1) and (3.2) in
Example 3.1. The rest of this example can be solved as before.
Examples 3.1 and 3.8 showed that we can use Cartesian coordinates (differentiated
in the inertial frame) to find the equations of motion of the simple pendulum, but the
results can be messy. In fact, for the simple pendulum, the result was not simple at all.
This motivates us to look for alternate coordinates to solve this and other problems.
Kinematics in Polar Coordinates
Of course Cartesian coordinates are not the only ones available for describing the
position of a point in the plane. The complexity of the pendulum example is a
strong motivator, for instance, for experimenting with different coordinates. Eq. (3.9)
showed how to represent a vector in polar coordinates. We can thus easily find the
inertial velocity and acceleration in terms of polar coordinates using Definition 3.3
(as was done in Example 3.6). By using the scalar product rule and chain rule on each
component, we have:
I
vP/O = (ṙ cos θ − r θ̇ sin θ)e1 + (ṙ sin θ + r θ̇ cos θ)e2
(3.18)
I
aP/O = (r̈ cos θ − 2ṙ θ̇ sin θ − r θ̈ sin θ − r θ̇ 2 cos θ)e1
+ (r̈ sin θ + 2ṙ θ̇ cos θ + r θ̈ cos θ − r θ̇ 2 sin θ)e2.
(3.19)
These equations look much more complicated than their counterparts in Cartesian
coordinates. Have we really gained anything by moving to polar coordinates? For
certain problems, yes! Let’s take another look at the simple pendulum, now using
polar coordinates.
64
CHAPTER THREE
O
g
O
θ
e2
–F sin θe2 – F cos θe1
r=l
e1
P
mPge1
I
(a) Coordinates
(b) Reference frame
(c) Free-body diagram
Figure 3.14 Solving the pendulum using polar coordinates in the inertial frame.
Example 3.9 Solving the Pendulum Using Polar Coordinates
in the Inertial Frame
As shown in Example 3.4, the simple pendulum is a single-degree-of-freedom system.
We should thus be able to describe its motion with only one coordinate rather than the
two Cartesian coordinates and a constraint. Here we explore using polar coordinates
to solve it. Figure 3.14 shows that the pendulum mass position can be defined using
the polar coordinates (r, θ )I , where I = (O, e1, e2, e3).
In polar coordinates, the position of the pendulum expressed as components in the
inertial frame is
rP/O = r cos θe1 + r sin θe2.
However, the constraint imposed by the fixed length of the pendulum rod is simply
r = l, implying ṙ = l˙ = 0. This eliminates the r coordinate—leaving only θ to describe
the one degree of freedom—part of our motivation for moving to polar coordinates
in this example. Taking the time derivative with respect to the inertial frame of the
position gives
I
vP/O =
Id
dt
(l cos θe1 + l sin θe2) = −l θ̇ sin θe1 + l θ̇ cos θe2.
Taking the time derivative again yields the inertial acceleration:
I
aP/O = (−l θ̈ sin θ − l θ̇ 2 cos θ)e1 + (l θ̈ cos θ − l θ̇ 2 sin θ)e2.
(3.20)
Using Eq. (3.20) and the free-body diagram in Figure 3.14c, we can write down
Newton’s second law for the pendulum bob:
−F sin θe2 − F cos θe1 + mP ge1 = mP (−l θ̈ sin θ − l θ̇ 2 cos θ)e1
+ mP (l θ̈ cos θ − l θ̇ 2 sin θ)e2.
This vector equation is equivalent to the two scalar equations,
−F cos θ + mP g = mP (−l θ̈ sin θ − l θ̇ 2 cos θ)
−F sin θ = mP (l θ̈ cos θ − l θ̇ 2 sin θ).
(3.21)
(3.22)
PLANAR KINEMATICS AND KINETICS OF A PARTICLE
65
As mentioned, part of our motivation for moving to polar coordinates is its compact
description of this single-degree-of-freedom system using the polar angle θ. We find
the equation of motion for θ by multiplying Eq. (3.21) by −sin θ and Eq. (3.22) by
cos θ and then adding the result to obtain
θ̈ = −
g
sin θ.
l
(3.23)
This is the equation of motion for coordinate θ describing the single degree of
freedom. It is a far simpler equation of motion than the two expressions in Eq. (3.7)
plus the constraint equation.
We can also easily solve for the constraint force in the rod by instead multiplying
Eq. (3.21) by cos θ and Eq. (3.22) by sin θ and adding to obtain
F = mP (g cos θ + l θ̇ 2).
Having this expression for the constraint force in the rod is useful if we want to
engineer a pendulum that won’t break. As mentioned in Chapter 1, we will often
solve for the constraint forces as well as the equations of motion. Remember, any
time there is a constraint that reduces the number of degrees of freedom, there is an
accompanying constraint force.
We can also solve for the equilibrium points of the simple pendulum by setting
θ̈ = 0 in Eq. (3.23):
0=−
g
sin θ.
l
There are two equilibrium angles for the pendulum: θ0 = 0 (the pendulum hanging
down) and θ0 = ±π (the pendulum standing up). As shown later in the book, the
inverted pendulum is an example of a system with an unstable equilibrium point: a
slight perturbation will cause the pendulum to fall.
Eq. (3.23) is a classic equation that appears often in dynamics. Of special significance is an approximation of the equation when the angle θ is small (i.e., for small
motion about the equlibrium θ0 = 0). In this case, we replace sin θ by its Taylor series
about θ0 = 0 and keep only the first-order term:4
sin θ = θ −
θ3 θ5 . . .
+
≈ θ.
+
5!
3!
(This approximation is discussed in Appendix A.)
After making this approximation, the pendulum equation becomes
g
θ̈ = − θ.
l
(3.24)
√
Eq. (3.24) is a simple harmonic motion equation with natural frequency ω0 = g/ l
(see Tutorial 2.3). Therefore, as long as θ is small, the pendulum mass oscillates
sinusoidally about θ0 = 0 and the frequency of oscillation is inversely proportional to
the square root of the length of the pendulum rod.
4 Also
called the small-angle approximation.
66
CHAPTER THREE
Not only were polar coordinates algebraically less cumbersome than Cartesian
coordinates for the pendulum, but the final solution was also much more elegant and
better behaved numerically. We have a single equation of motion for the degree of
freedom represented by θ and we eliminated the constraint force in the rod. However,
the algebra was still involved—we had to solve for two scalar equations and then
manipulate these equations to find the equation of motion. Soon we will introduce
another concept—the polar frame—that will make the pendulum problem even easier
to solve.
Kinematics in Path Coordinates
Path coordinates don’t lend themselves easily to differentiation in the inertial frame
because there is no representation for the position of P with respect to the origin O
of the inertial frame. The scalar coordinate s is not sufficient to specify the (planar)
position of P without a functional form for the path. In fact, path coordinates are
rarely used in cases where the position of a particle is of interest. Typically, it is only
the particle motion (and the accompanying forces) we care about. Thus finding the
acceleration and applying Newton’s second law using path coordinates requires more
sophisticated approaches that we delve into next.
3.4 Inertial Velocity and Acceleration in a Rotating Frame
If we were only interested in representing the position and velocity of a particle in
terms of components in an inertial frame and in solving for its trajectory there, we
would be done. We could end this chapter here and move on to other topics. But then
why do we place so much emphasis on the reference-frame concept? The answer is
that it is unnecessarily limiting and complicated to solve every problem using vectors
expressed as components in an inertial frame. Many, many problems are more easily
solved—or produce expressions that are easier to work with—when we introduce
additional, noninertial reference frames. This is the subject of the remainder of this
chapter.
Introducing one or more additional reference frames to a problem offers several
advantages and will become a fundamental piece of our problem-solving machinery
throughout the book. First, it is often important for engineers to build models that
relate to actual measurements. Since these measurements are often obtained in moving or rotating frames, it is very convenient to have expressions for the dynamical
trajectories in terms of coordinates in those frames. Even measurements from the
ground, tracking an airplane, say, are in a moving and rotating frame—fixed to the
earth—as opposed to an inertial frame fixed to the distant stars; see Figure 3.15. Second, many problems have constraints that reduce the number of degrees of freedom.
For these problems, operating in absolute space often means carrying around more
scalar coordinates than necessary, and thus we have to explicitly utilize the constraint
equations (see Example 3.1). We can often simplify such problems by introducing a
new reference frame. Finally, moving and rotating frames are an essential construct
for keeping track of the various coordinates associated with multiple particles and/or
PLANAR KINEMATICS AND KINETICS OF A PARTICLE
67
Sun
Earth
Moon
Figure 3.15 Illustration of multiple reference frames.
multiple rigid bodies. The solution to almost every dynamics problem begins with the
careful and judicious choice of appropriate and useful reference frames.
In this section we discuss one of the key uses of rotating frames: to simplify finding
the velocity and acceleration in different coordinates. For instance, while procedurally
deriving Eqs. (3.18) and (3.19) was reasonably straightforward (though algebraically
messy), the resulting expressions for the velocity and acceleration are rather involved.
Introducing a new frame and expressing the inertial velocity and acceleration in it
greatly reduces our work.
3.4.1 Relative Orientation
Before discussing the use of alternate reference frames in forming the velocity and
acceleration, we must introduce the geometric tool used to describe two reference
frames that differ in orientation. What, in fact, do we mean by the orientation of one
frame relative to another? For now, we only consider planar orientation. That is, we
consider only the relative orientation of two reference frames that share a common
axis (typically, the third axis). Chapter 10 presents the more general description of
three-dimensional relative orientation.
Figure 3.16 shows two reference frames, A = (O, a1, a2, a3) and B =
(O, b1, b2, b3), where a3 = b3. Frame B is rotated with respect to frame A by an
angle θ about their common axis. Since b1 is a vector, it can be expressed as components in A:
b1 = cos θa1 + sin θa2.
Likewise,
b2 = − sin θa1 + cos θa2.
We call such a change in orientation about an axis fixed in both frames a simple
rotation.
68
CHAPTER THREE
A
B
a2
b2
b1
θ
O
a1
Figure 3.16 Reference frame B = (O, b1, b2 , b3) is rotated with respect to frame A =
(O, a1, a2 , a3) by angle θ.
These relationships can be compactly represented by writing them as rows in a
vector transformation table:
a1
b1
b2
cos θ
− sin θ
a2
sin θ
cos θ ,
where the row i and column j element is given by bi . aj .
The transformation-table notation also makes it straightforward to write the A unit
vectors in terms of the B unit vectors by summing each column of the table:
a1 = cos θb1 − sin θb2
a2 = sin θb1 + cos θb2.
So, for example, given a vector rP/O expressed as components in A (i.e., rP/O =
a1a1 + a2a2), we can use the transformation table to express the vector in terms of
components in B,
rP/O = a1(cos θb1 − sin θb2) + a2(sin θb1 + cos θb2)
= (a1 cos θ + a2 sin θ) b1 + (−a1 sin θ + a2 cos θ) b2,
=b2
=b1
where (b1, b2)B are the scalar magnitudes of the components of rP/O expressed in
frame B.
We can also use matrix notation to write the transformation of the component
magnitudes in matrix form:
b1
cos θ
sin θ
a1
=
,
b2 B
a2 A
− sin θ cos θ
=BC A
where BC A is the transformation matrix from frame A to frame B. Its elements
are likewise given by Cij = bi . aj . The inverse transformation is given by AC B =
(BC A )−1 = (BC A )T .
We use the transformation table throughout the book to describe the relative
orientation of one frame with respect to another. The important thing to remember is
PLANAR KINEMATICS AND KINETICS OF A PARTICLE
69
that when presented with a vector, we have a variety of frame choices for writing the
components. Expressing the components of a vector (particularly the acceleration)
in one frame versus another can make problems significantly easier to solve. The
transformation table introduced in this section is used to transform the components
of a vector from the unit-vector directions of one frame to the unit-vector directions
of another.
Note that Figure 3.16 may give the misleading impression that the transformation
table only applies when the two frames share a common origin. This is not the case.
Remember, the unit vectors are just vectors of unit magnitude. It does not matter
where they are located. We can translate frame B anywhere and, as long as its relative
orientation to I stays the same, the transformation table between the two sets of unit
vectors is the same.
Example 3.10 Satellite Tracking, Revisited
Example 3.3 showed how to express a vector as components in two different frames.
We repeat that example here using our new formalism for relative orientation. Recall
that Example 3.3 defined two frames of reference: an inertial frame I located at the
center of the earth O and a second frame B located on the surface of the earth at origin
O and sharing the e1 = b1 axis with frame I . Even though the two frames do not share
the same origin, the unit vectors defining them are still related by a transformation
table using the fact that B is rotated by −45◦ about e1 from I :
b2
e2
e3
cos 45◦
− sin 45◦
b3
sin 45◦
cos 45◦ .
The transformation table shows that we can write the b2 unit vector in terms of the e2
and e3 unit vectors as
1
1
b2 = cos 45◦e2 − sin 45◦e3 = √ e2 − √ e3,
2
2
which is the same as Eq. (3.8) in Example 3.3. The rest of the example follows as
before.
3.4.2 Two Rotating Frames
By now you should be convinced that finding equations of motion for a particle
using Newton’s second law requires expressing the acceleration relative to an inertial
frame. This is fundamental. Earlier in the chapter we showed how to find the velocity
and acceleration in terms of both Cartesian and polar coordinates (we postponed
path coordinates) and how to use the resulting vector expressions to find equations
of motion. This main point cannot be made often enough: to compute the inertial
kinematics for use in Newton’s second law, the vector derivative must be taken with
respect to an inertial frame.
However, we are free to express the resulting velocity and acceleration as components in any frame we like. As noted for Eqs. (3.18) and (3.19), expressing them
as components in the inertial frame can result in rather complicated and lengthy expressions. In the next two subsections we introduce two important reference frames.
70
CHAPTER THREE
B
I
e2
eθ
P
r
er
θ
O
e1
Figure 3.17 Polar coordinates (r, θ )I and polar frame B = (O, er , eθ , e3), where e3 = er × eθ
points out of the page (not shown).
Either can be used for expressing the components of IvP/O and IaP/O and, for certain
problems, these representations greatly simplify finding the motion. These frames can
also be used to find a new and elegant approach for calculating the inertial kinematics.
The Polar Frame
The polar frame is a frame of reference that shares an axis with the inertial frame
(usually the e3 or ez axis) and has a unit vector defining a second axis (usually the
er axis) directed from the frame’s origin to a point P even as P moves. In the case
where the polar frame B has the same origin O as the inertial frame I , it is particularly
simple to express the position of P with respect to O as components in the polar frame
using the polar radius r:
rP/O = rer .
(3.25)
The unit vector eθ points in the direction of increasing θ . Since er always points at P ,
eθ is not needed to describe the position of P relative to O as components in B.
Although both the polar radius r and polar angle θ are needed to locate a point in
the inertial frame, we can locate P in the polar frame using only r. Of course, we can’t
ignore θ altogether, as it defines the orientation of the polar frame with respect to the
inertial frame (see Section 3.4.1). The polar angle is thus used to express the position
rP/O as components in the inertial frame. The polar frame is particularly convenient
for constrained problems that have only a single degree of freedom, such as the simple
pendulum that we study extensively in this chapter.
The polar frame is completed by eθ = e3 × er , which is perpendicular to er in
a right-hand sense. (See Appendix B for properties of the vector cross product.)
Figure 3.17 shows the polar frame B = (O, er , eθ , e3). As P moves on some arbitrary
path, frame B rotates relative to frame I and is thus not necessarily inertial, unless θ
is constant.
The real value of introducing the polar frame is that the inertial acceleration of P
in Newton’s second law can be expressed in terms of the unit vectors er and eθ : this
representation can simplify—and enhance our understanding of—certain problems.
We start by writing the kinematics—that is, the inertial velocity and acceleration—
as components in the polar frame. In Eqs. (3.18) and (3.19), these kinematics were
written in polar coordinates but as vector components in the inertial frame. Using
the transformation table in Section 3.4.1, we can express the inertial unit vectors as
components in the polar frame:
PLANAR KINEMATICS AND KINETICS OF A PARTICLE
71
e1 = cos θer − sin θeθ
e2 = sin θer + cos θeθ .
Substituting these expressions into Eqs. (3.18) and (3.19), we obtain (after some
algebra)
I
(3.26)
I
(3.27)
vP/O = ṙer + r θ̇eθ
aP/O = (r̈ − r θ̇ 2)er + (2ṙ θ̇ + r θ̈)eθ .
Eqs. (3.26) and (3.27) are important equations and are much simpler than
Eqs. (3.18) and (3.19). These equations also have important physical consequences
that are discussed in Section 3.5. It is important to remember that the velocity and
acceleration vectors given in Eqs. (3.18) and (3.19) and the vectors in Eqs. (3.26)
and (3.27) are the same. We have simply expressed them as vector components in
two different reference frames.
The process by which we obtained the kinematics in the polar frame, however, was
a bit involved. And this process doesn’t always work out so cleanly, as discussed in
Section 3.3 with regard to path coordinates. It would be much more useful to directly
find the inertial derivative of a vector expressed as vector components in the rotating
frame. This alternative is conceptually challenging, however, because the unit vector
associated with each component may be changing with time (recall that the polar
frame is rotating with respect to the inertial frame). The advantage, though, is having a
procedure that provides a completely general formalism for finding the time derivative
of a vector expressed in a rotating frame. We do this next for the polar frame. Chapter 8
generalizes this approach to any (planar) rotating frame.
Starting with Eq. (3.25) for rP/O in frame B, the inertial velocity is written as
I
vP/O =
Id
dt
rP/O =
Id
dt
(rer ) = ṙer + r
Id
dt
(er ),
where we have used the product rule of vector derivatives.
I
What is dtd (er )? If we write the unit vector er as
er = cos θe1 + sin θe2,
then we can use Definition 3.3 to find its derivative in I :
Id
dt
(er ) = −θ̇ sin θe1 + θ̇ cos θe2
= θ̇(− sin θe1 + cos θe2).
Using the vector transformation table in Section 3.4.1 to convert back to the polar
frame, we find
Id
dt
I B
(er ) = θ̇eθ =
ω eθ .
(3.28)
72
CHAPTER THREE
Here we introduced an important new symbol, IωB =
θ̇, to represent the time rate of
change of the orientation of frame B in frame I . The inertial velocity of P expressed
as components in B is
I
vP/O = ṙer + r θ̇ eθ ,
(3.29)
which is the same as the velocity given in Eq. (3.26).
We can follow the same procedure to find the derivative of the eθ unit vector:
Id
dt
−IωB er .
(eθ ) = −θ̇er =
(3.30)
Eqs. (3.28) and (3.30) can be written more compactly by introducing a vector
quantity—the angular velocity—defined as
I B I B
ω = ω e3,
(3.31)
where e3 is a unit vector mutually perpendicular to er and eθ , forming a right-hand set.
We call an angular velocity with a fixed inertial direction a simple angular velocity.
Definition 3.4 A simple angular velocity is a vector aligned with the common,
fixed axis of rotation when a reference frame rotates relative to another reference
frame about a direction fixed in both frames. Typically that axis is one of the unit
vector directions of both frames. The magnitude of the simple angular velocity
is the rate of rotation about the axis.
The angular velocity points along the axis of rotation of frame B. Its magnitude
is the instantaneous rate of rotation about that axis. Because we are only treating
planar motion in this chapter, the angular velocity takes a particularly simple form—
it is always directed along the axis perpendicular to the plane containing the particle
motion. We postpone until Chapter 10 the more general three-dimensional treatment.
Inspection of Eqs. (3.28) and (3.30) reveals that the angular velocity in Eq. (3.31)
can be used to write the following compact expressions for the inertial derivative of
the unit vectors of the polar frame:
Id
dt (er )
=
IωB
× er
Id
=
IωB
× eθ .
dt (eθ )
(3.32)
You can check these formulas by substituting Eq. (3.31) into Eq. (3.32) and comparing
the result to Eqs. (3.28) and (3.30). The formulas in Eq. (3.32) are extremely important
and are used throughout the book. MEMORIZE THEM!
Eq. (3.32) is true for any unit vector fixed in a frame B that is rotating in another
frame I ,
Id
dt
b = IωB × b,
where b is a unit vector fixed in B. One helpful way to remember this formula is that
the superscript to the left of ω is the same as the superscript on the derivative to the
PLANAR KINEMATICS AND KINETICS OF A PARTICLE
eθ
O
g
O
θ
r=l
73
B
e2
–Fer
er
e1
mPge1
I
P
(a) Coordinates
(b) Reference frames
(c) Free-body diagram
Figure 3.18 Solving the pendulum using polar coordinates in the polar frame.
left; it is the frame in which B is rotating. The superscript to the right of ω is the frame
in which the unit vector is fixed (not rotating).
Using Eqs. (3.29) and (3.32), the acceleration of point P can be computed with
respect to reference frame I and expressed as components in the polar frame much
more easily than before (see if you can do this):
I
aP/O = (r̈ − r θ̇ 2)er + (2ṙ θ̇ + r θ̈)eθ .
(3.33)
Again, this result is the same as in Eq. (3.27), but we obtained it much more directly.
Remember that we are computing the time derivative of a vector with respect to
a reference frame. In this case, it is the rate of change of the vector with respect to
the inertial frame. The only difference between Eqs. (3.18) and (3.19) and Eqs. (3.29)
and (3.33) is the frame in which we have expressed the components of the respective
vectors. Although this distinction may seem minor, it makes solving many problems
much more straightforward. To demonstrate, we solve for the equations of motion of
the simple pendulum once again.
Example 3.11 Solving the Pendulum Using Polar Coordinates
and the Polar Frame
The polar frame B illustrated in Figure 3.18b has angular velocity
I B
ω = θ̇e3.
We relate frame B to the inertial frame I by using the polar angle θ to generate the
transformation table
er
e1
e2
cos θ
sin θ
eθ
− sin θ
cos θ .
We use this table to convert each unit vector from one reference frame to the other:
e1 = cos θer − sin θeθ
(3.34)
e2 = sin θer + cos θeθ
(3.35)
74
CHAPTER THREE
and
er = cos θe1 + sin θe2
(3.36)
eθ = − sin θe1 + cos θe2.
(3.37)
The position rP/O of the pendulum bob P with respect to O is
rP/O = ler .
(3.38)
Taking the time derivative with respect to the inertial frame of the position in Eq. (3.38)
and using Eq. (3.32) gives the inertial velocity of P expressed as vector components
in the polar frame:
I
vP/O =
Id
dt
(ler ) = l IωB × er = l θ̇eθ .
(3.39)
Similarly, taking the time derivative with respect to the inertial frame of the velocity
in Eq. (3.39), using Eq. (3.32) and the product rule, gives the inertial acceleration
of P ,
I
aP/O =
Id
dt
(l θ̇eθ ) =
d
(l θ̇)eθ + l θ̇ IωB × eθ = l θ̈eθ − l θ̇ 2er .
dt
(3.40)
Next, using Eq. (3.40) and the free-body diagram in Figure 3.18c, we write down
Newton’s second law for the pendulum bob:
−F er + mP ge1 = mP (l θ̈eθ − l θ̇ 2er ).
(3.41)
To derive the equations of motion, it is easiest—in this case—to equate vector
components in the polar frame. We thus need to write Eq. (3.41) entirely as vector
components in the polar frame. Using Eqs. (3.34) and (3.41), we obtain
−F er + mP g(cos θer − sin θeθ ) = mP (l θ̈eθ − l θ̇ 2er ).
(3.42)
Eq. (3.42) is equivalent to the two scalar equations
−F + mP g cos θ = −mP l θ̇ 2
−mP g sin θ = mP l θ̈.
(3.43)
(3.44)
Simplifying Eq. (3.44), we once again obtain the equation of motion for the pendulum
bob in polar coordinates (see Eq. (3.23)),
θ̈ = −
g
sin θ.
l
As in Example 3.9, we can also solve for the force in the rod using Eq. (3.43) to
verify that the pendulum can support this constraint force.
Example 3.12 A Particle on a Spring Free to Move in the Plane
Consider a simple two-degree-of-freedom problem in which a particle of mass mP
connected to the origin via a linear spring is free to move anywhere in the plane
PLANAR KINEMATICS AND KINETICS OF A PARTICLE
I
B
I
eθ
r
e2
P
P
75
er
k θ
O
O
–k(r – r0)er
e1
(b) Reference frames
(a) Coordinates
(c) Free-body diagram
Figure 3.19 A two-degree-of-freedom problem consisting of a mass on a spring free to move
in the plane.
(Figure 3.19). From the free-body diagram (Figure 3.19c), we see that the spring
applies a force only when it is stretched radially. Thus polar coordinates are a sensible
choice to describe the particle’s position in the plane. By using a polar frame, the force
on the particle is easily written as
FP = −k(r − r0)er ,
where k is the spring constant and r0 is the unstretched length of the spring. The
acceleration of the particle in polar coordinates using the polar frame is given by
Eq. (3.33). Newton’s second law then becomes
I
aP/O = (r̈ − r θ̇ 2)er + (2ṙ θ̇ + r θ̈)eθ = −
k
(r − r0)er .
mP
(3.45)
Eq. (3.45) is equivalent to the two scalar equations
r̈ +
k
(r − r0) − r θ̇ 2 = 0
mP
(3.46)
ṙ
θ̈ + 2 θ̇ = 0.
r
(3.47)
These two scalar equations can be integrated numerically for various initial conditions
to find the trajectory of the pendulum described by (r, θ )I .
Two special analytical solutions stand out. If the particle is given no initial speed
in the eθ direction, so that θ̇(0) = 0, then θ̈ = 0 for all time and the particle undergoes
simple harmonic motion in the radial direction with angle θ(0); it is the same solution
as the important one-degree-of-freedom problem studied in Tutorial 2.3.
There is a second interesting solution with θ̇ = constant for all time: the particle
travels in a circle at constant angular velocity. Setting r̈ = ṙ = θ̈ = 0 in Eqs. (3.46)
and (3.47), the condition for circular motion is
k
(r − r0) = r θ̇(0)2.
mP
76
CHAPTER THREE
This equation can be solved for the steady-state radial offset of the particle,
rss = k
m P r0
k
mP
− θ̇(0)2
.
Finally, the equations of motion (Eqs. (3.46) and (3.47)) can be simplified by
recognizing that Eq. (3.47) is a perfect differential,
d 2 r θ̇ = 0,
dt
or
r 2θ̇ = hO ,
(3.48)
where hO is some constant depending on initial conditions. Eq. (3.48) lets us rewrite
Eq. (3.46) as
r̈ +
h2
k
(r − r0) − O3 = 0.
mP
r
Making this observation reduces the two-degree-of-freedom problem to a single
differential equation in r, thus simplifying the analysis. Once the radial trajectory
is found, the angular trajectory θ(t) is found from Eq. (3.48). Such situations are
further discussed in Chapter 4.
The Path Frame
A rotating frame is especially helpful when finding the kinematics of a point using
path coordinates. Path coordinates are normally used when we are not interested in
the location of a point relative to an origin but instead want to know how its velocity
and acceleration vary along a predefined trajectory (e.g., to determine whether the
acceleration of a car traveling on a curved track leads to skidding). This motivates us
to define the path frame, whose origin is the point itself.
We define the path frame as B = (P , et , en, e3). The tangent unit vector et is
directed along the (instantaneous) velocity IvP/O and is thus always tangent to the
path. The normal unit vector en is perpendicular (normal) to the path (Figure 3.20).
We use the convention that en always points inward from the path curvature. That is,
the unit vector e3 = et × en points out of the plane of the page when the path turns
counterclockwise, and it points into that plane when the path turns clockwise. In a
moment we will see how this convention is established by the instantaneous angular
velocity IωB .
The inertial velocity of point P written as components in B is
I
vP/O = ṡet =
vet .
Note that we have kept the O in the subscript here, to represent the origin of the frame
in which the path is defined.
PLANAR KINEMATICS AND KINETICS OF A PARTICLE
I
P vP
77
et
ey
s
en
O
B
ex
Figure 3.20 Path coordinate s measures the position of point P along an arbitrary path in
inertial frame I . The path frame is B = (P , et , en, e3), where e3 is into the page at the instant
shown.
The acceleration is then found the usual way by differentiating the velocity with
respect to the inertial frame. We obtain
I
aP/O = v̇et + v
Id
dt
(et ).
(3.49)
I
We are again faced with having to find the derivative of a unit vector. To find dtd (et ),
observe that the path frame B behaves just like the polar frame—or any frame that
rotates with respect to I . That is, as the point P moves a small amount ds in a small
time dt, the frame B rotates a small amount dθ due to the curvature of the path. At that
instant, the frame is rotating with respect to I with angular velocity IωB =
θ̇e3 = dθ
dt e3.
The time rate of change of the unit vector et is thus given by Eq. (3.32) with er replaced
by et :
Id
dt
(et ) = IωB × et = θ̇en.
As mentioned above, the direction of en depends on the path curvature. But what
is the relationship between the path curvature and θ̇? We find this relationship by
approximating the path as a series of circular segments. An infinitesimal displacement
ds along the path can be modeled by an infinitesimal arc of a circle, as illustrated in
Figure 3.21. The angle subtended by the circular arc ds in time dt is
dθ =
ds
,
Rc
where Rc is the radius of curvature of the path at that point.
Dividing by dt, we find
1 ds
v
dθ
= θ̇ =
=
,
dt
Rc dt
Rc
ṡ is the (instantaneous) speed along the path. Remember that Rc is not a
where v =
constant but varies along the path as the curvature changes—it is a property of the
path geometry. Substituting this result into Eq. (3.49) allows us to write the inertial
78
CHAPTER THREE
ds
Rc
dθ
Figure 3.21 The radius of curvature of a small segment ds of the path.
acceleration of P in terms of its instantaneous speed along the path and the local
radius of curvature:
I
aP/O = v̇et +
v2
en .
Rc
The final result needed is a relationship between the radius of curvature and the
path geometry. If the path can be defined by a function y(x) relating the Cartesian
coordinates x and y, then the radius of curvature of the path is
1+
Rc =
dy
dx
d 2y
dx 2
2 23
.
(3.50)
You have an opportunity to derive this relationship in Problem 3.13.
Example 3.13 Charged Particle in a Magnetic Field
This example uses the path frame B = (P , et , en, e3) to find the equations of motion of
a charged particle in a uniform and constant magnetic field (Figure 3.22). Let q = 0
denote the electrical charge of particle P . Orient the reference frame such that the
magnetic field is given by Bez ,5 where B > 0 and I = (O, ex , ey , ez ) is an inertial
frame. We define B so that e3 = ez if the particle is turning counterclockwise, and e3 =
−ez if the particle is turning clockwise. According to the laws of electromagnetism,
the force on the particle is given by the Lorentz force
FP = q IvP/O × Bez .
The kinematics of P in terms of vector components in B are
I
vP/O = ṡet
I
aP/O = s̈et +
5A
ṡ 2
en .
Rc
magnetic field is a vector field, that is, at each point in space, the local magnetism has both magnitude
and direction.
PLANAR KINEMATICS AND KINETICS OF A PARTICLE
B
79
I
ez
ey
P
s
et
ex
P
e3
(a) Coordinates
B
en
(b) Reference frames
±qs·Ben
(c) Free-body diagram
Figure 3.22 Charged particle in a magnetic field.
Using Newton’s second law, we have
ṡ 2
en = q ṡet × (±Be3) = ±q ṡBen.
mP s̈et +
Rc
We obtain the two scalar equations
s̈ = 0
ṡ = ±
qBRc
.
mP
The first equation—the equation of motion—reveals that the speed of the particle
is constant: ṡ(t) = ṡ(0) > 0. Using the second equation, we compute the radius of
curvature
Rc =
mP
ṡ(0) > 0,
|q|B
where, by convention, we adopt the positive solution. Thus a charged particle in a
magnetic field travels in a circle in a plane perpendicular to the direction of the field.
The radius of the circle is proportional to the mass and initial speed of the particle and
is inversely proportional to the magnitude of the electrical charge and the magnitude
of the magnetic field. The direction of rotation depends on the charge of the particle.
For example, a negatively charged particle rotates in the direction of the fingers of
your right hand when you curl them so that your thumb points along the magnetic
field vector Bez .
3.5 The Polar Frame and Fictional Forces
It is worthwhile to pause at this point to discuss in more depth the formula for
acceleration in Eq. (3.33) (repeated here for convenience):
I
aP/O = (r̈ − r θ̇ 2)er + (2ṙ θ̇ + r θ̈)eθ .
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CHAPTER THREE
P at t + Δt
I
I
e2
vP/O
P at t
B
eθ
θ(t + Δt)
er
θ(t)
O
e1
Figure 3.23 Point P traveling in a straight line in absolute space with velocity IvP/O and the
rotating polar frame B.
This expression often causes great confusion. When expressed in polar coordinates
and the polar frame, the acceleration is no longer a simple second-derivative of a
scalar coordinate—it includes two additional terms, one in each of the polar-frame
directions. It is helpful to discuss the physical meaning of Eq. (3.33) and how it results
from the geometry of motion.
First, remember that the acceleration IaP/O in Eq. (3.33) is still the acceleration
of point P with respect to absolute space. We have just used polar coordinates rather
than Cartesian coordinates and expressed it as vector components in the polar frame.
Using polar coordinates and the polar frame does not change the fundamental nature
of the acceleration. Consider, for example, the simple situation of a point P traveling
at a constant velocity in the inertial frame I with origin O, as shown in Figure 3.23.
The inertial acceleration of P is zero, of course, and its velocity could be written quite
simply in Cartesian coordinates as ẋe1 + ẏe2, where (ẋ, ẏ)I are both constant. The
second derivatives of the Cartesian coordinates are zero.
Now use polar coordinates (r, θ )I to locate P and add a polar frame B =
(O, er , eθ , e3), where the er unit vector points at P (Figure 3.23). As P travels along
a constant-velocity straight-line path in absolute space, its radial position r increases,
and B must rotate for er to continue pointing at P . Since the inertial acceleration is
zero, Newton’s second law becomes
FP = 0 = mP (r̈ − r θ̇ 2)er + mP (2ṙ θ̇ + r θ̈)eθ .
(3.51)
Since the magnitude of each vector component in Eq. (3.51) equals zero, the second
derivatives r̈ and θ̈ of the polar coordinates are not zero!
We call the acceleration-like term −r θ̇ 2er the centripetal acceleration and 2ṙ θ̇eθ
the Coriolis acceleration. For instance, Eq. (3.51) can be written as the two scalar
equations
r̈ = r θ̇ 2
θ̈ = −2
ṙ θ̇
.
r
PLANAR KINEMATICS AND KINETICS OF A PARTICLE
81
These equations look a lot like Newton’s second law for polar coordinates. As a result,
the terms on the right are frequently called fictional forces. We want to emphasize that
these are kinematic terms and not forces. No forces are actually acting; they just look
like forces from the way they appear in these equations.
What is going on here? Cartesian coordinates really are special when it comes to
dynamics. Only in Cartesian coordinates do we expect to see simple expressions of
Newton’s second law in terms of the second derivative of each coordinate. In other
coordinate systems, the scalar derivatives are coupled and there may be extra terms.
The appearance of these terms is a purely kinematic effect.
In Problem 3.9 you have the opportunity to explore this simple example a bit
further, solving for r̈ and θ̈. You will find that the polar angle θ approaches an
asymptotic value as the radial unit vector er gradually aligns with the particle’s path.
In the remainder of this section we further explore the kinematic source of these
two terms in the accelerations.
3.5.1 Centripetal Acceleration
Consider next a slightly different scenario. Here point P is traveling on a circle of
radius r at angular rate θ̇, as shown in Figure 3.24. In other words, the position of
P is fixed in a rotating polar frame B = (O, er , eθ , e3) whose origin O is at the
center of the circle. The fact that the radius r is constant implies r̈ = ṙ = 0. The
velocity IvP/O at time t is tangent to the circle and has magnitude r θ̇(t). At some
short time t later, the velocity, while still of the same magnitude, has changed by
I
vP/O (t + t) − IvP/O (t). Any time the velocity changes
a small amount IvP/O =
magnitude or direction relative to a particular frame, there must be, by definition,
an acceleration relative to that frame. The inertial acceleration IvP/O /t accounts
for the change in direction of the velocity. This acceleration is exactly equal to the
centripetal acceleration −r θ̇ 2er , which points inward toward O.
I
I
B
vP/O(t + Δt)
I
vP/O(t)
e2
P
eθ
I
ΔvP/O
er
O
I
vP/O(t + Δt)
I
vP/O(t)
e1
(a)
(b)
Figure 3.24 (a) Point P is traveling in a circle in the inertial frame I . Polar frame B stays
aligned with the vector toward P . (b) Over a short time interval the velocity changes by an
I
amount IvP/O =
vP/O (t + t) − IvP/O (t) in the radial direction, resulting in centripetal
acceleration.
82
CHAPTER THREE
O
(a)
P
O
P
(b)
Figure 3.25 Particle trajectories in the inertial frame. (a) Centripetal force with ṙ(0) = 0 yields
circular motion in the inertial frame. (b) Centripetal and Coriolis forces with ṙ(0) < 0 yield
spiral motion in the inertial frame.
Since the velocity of P is changing as it travels in a circle, there must be a force
accelerating P with respect to absolute space. From Eq. (3.33), that force satisfies
FP . er = mP (r̈ − r θ̇ 2) = −mP r θ̇ 2.
(3.52)
The radial force component in Eq. (3.52) is called a centripetal force. The circular trajectory of a particle acted on by a constant centripetal force is shown in Figure 3.25a.
3.5.2 Coriolis Acceleration
Now consider particle P traveling around O at constant angular rate θ̇ and moving
radially inward at constant rate ṙ < 0. In the polar frame, the particle will move in
a straight line at constant speed toward the origin. In absolute space, however, the
particle trajectory is a spiral, as shown in Figure 3.25b. For the particle to follow
such a trajectory, clearly a force must be acting. Newton’s second law tells us that an
additional force must be acting in the eθ direction:
FP . eθ = mP (2ṙ θ̇ + r θ̈) = 2mP ṙ θ̇.
(3.53)
The tangential force component in Eq. (3.53) is called the Coriolis force.
The implications of the Coriolis force are more subtle than those of the centripetal
force and thus tend to cause more confusion. Physically, the Coriolis acceleration
arises from two sources (hence the factor of 2 in Eq. (3.33)). The first source comes
from the fact that, as the point P moves inward (outward), it must move slower
(faster) in the eθ direction relative to absolute space to continue rotating at a constant
angular rate θ̇. Thus over a small time period, its speed decreases (increases), hence
the acceleration. The second source arises because, over this small time interval, the
direction of the inertial velocity changes as the frame B rotates with respect to I .
Each source contributes a factor of ṙ θ̇ to the acceleration.
We revisit centripetal and Coriolis accelerations in Chapter 8 when discussing
motion relative to a rotating frame.
PLANAR KINEMATICS AND KINETICS OF A PARTICLE
83
3.6 An Introduction to Relative Motion
Another situation where we need to introduce additional frames of reference is when
describing the motion of a particle relative to a moving platform. Rather than use
scalar coordinates to specify the position of P in a fixed inertial frame, we locate the
particle in a moving frame B. This frame can be either translating or rotating. For
instance, many problems on the surface of the earth must account for the fact that a
reference frame fixed to the earth is both rotating and translating (because of its orbit
about the sun) in inertial space.
For most problems involving relative motion (e.g., an accelerating or rotating
frame), we cannot employ Newton’s second law in a moving frame, even though
we may want to solve for trajectories in that frame; remember, Newton’s second
law works only when the acceleration is relative to absolute space (i.e., an inertial
frame). We need to be more careful. This section introduces the problem of relative
motion by solving for the equations of motion of a particle relative to a reference frame
translating (but not rotating) in inertial space. One outcome is the observation that all
reference frames traveling at constant velocity are inertial. The more complicated
situation of a particle moving relative to a frame both translating and rotating is
deferred to Chapter 8.
Consider the situation depicted in Figure 3.26. Frame I = (O, e1, e2, e3) is a
fixed inertial frame while frame B = (O , ex , ey , ez ) is translating relative to I with
acceleration IaO /O . That is, we have
I
vO /O =
I
aO /O =
Id
dt
Id
dt
rO /O
vO /O .
I
Frame B is not rotating with respect to I .
Consider the motion of point P in Figure 3.26 with position rP/O in B. Using
Cartesian coordinates (x, y)B to describe the position of P in B, we have
rP/O = xex + yey
B
I
ey
e2
P
rP/O′
I
rP/O
rO′/O
O
O′
a O′/O
ex
e1
Figure 3.26 Translating reference frame B = (O , ex , ey , ez ) accelerates in I = (O, e1, e2 , e3)
by IaO /O .
84
CHAPTER THREE
or in matrix notation,
[rP/O ]B =
x
y
.
B
Our goal is to find the equations of motion for scalar coordinates (x, y)B describing
the trajectory of P in B (of course, any other scalar coordinates would work as well).
Since B may not be inertial, we cannot simply use Newton’s second law on BaP/O .
Instead we must first find the motion of P relative to I . Using vector geometry, the
position of P relative to O is
rP/O = rO /O + rP/O .
(3.54)
From the vector triad in Eq. (3.54) we have
I
vP/O =
Id
dt
rO /O +
Id
dt
rP/O = IvO /O + IvP/O .
(3.55)
Assume that IvO /O is known. What is IvP/O ? Using the expression for rP/O in
Cartesian coordinates in B yields
I
vP/O =
Id
dt
(xex + yey ).
We can now use the product rule on each component as before. However, if B is
only translating, the two unit vectors ex and ey do not change in I —there is no
angular velocity of B in I (remember Eq. (3.32)?). Since these are unit vectors, if
they don’t change their magnitude or direction in I , then they have no derivative in
I . Consequently, we find
I
vP/O = ẋex + ẏey ,
which implies, by the definition of the vector derivative,
I
vP/O = B vP/O .
This result may seem counterintuitive. How can the derivative of the position (the
velocity) be the same relative to either frame? Hopefully, a bit of thought makes
it clear. Translation of the vector rP/O does not change its magnitude or direction
in space. Since these are the only ways a vector can change, its derivative is the
same whether referred to B or I . Of course, this is only true for a translating frame.
Chapter 8 shows that things get more complicated in a rotating frame.
We now return to the original problem of finding the inertial acceleration of P in I
so that Newton’s second law can be applied. Take the derivative with respect to time
of Eq. (3.55) to obtain (using IaP/O = BaP/O )
aP/O = IaO /O + BaP/O = IaO /O + ẍex + ÿey .
I
(3.56)
PLANAR KINEMATICS AND KINETICS OF A PARTICLE
O
x
O′
O
e2
85
b2
O′
g
θ
e1
l
mP
a
P
I
–F cos θe1 – F sin θe2
b1
B
P
(a) Coordinates
(b) Reference frames
mPge1
(c) Free-body diagram
Figure 3.27 Pendulum in an accelerating box.
The acceleration of P relative to I is the sum of the acceleration of the origin of frame
B relative to I and that of P relative to B. We can use Eq. (3.56) to find the variant of
Newton’s second law that determines the motion of P in B. Multiplying Eq. (3.56)
by the mass mP and rearranging yields
mP BaP/O = FP − mP IaO /O ,
(3.57)
where we have substituted FP for mP IaP/O using Newton’s second law.
Thus to find the equations of motion for a particle relative to a translating frame,
we have to subtract the acceleration of the translating frame from the right side of
Newton’s second law. If frame B is moving at a constant velocity, then IaP/O = BaP/O and it makes no difference whether we write Newton’s second law in frame I or frame
B. This scenario is a demonstration of Newtonian relativity: Every nonrotating frame
moving at constant velocity in absolute space is an inertial frame.
Example 3.14 Pendulum in an Accelerating Box
Consider again the simple pendulum, but now let the attachment point be fixed to a
box that is accelerating in the inertial e2 direction (Figure 3.27). The goal is to find the
equation of motion for the pendulum relative to a frame fixed to the box. We will also
see that the steady-state equilibrium angle of the pendulum is offset from the vertical
when the box is accelerating.
Figure 3.27a shows the coordinates describing the motion. As in the stationary
pendulum, θ is the angle of the pendulum from the vertical. Let x denote the position
of the attachment point O with respect to origin O of the inertial frame. The velocity
ae2, respectively. We wish to find
and acceleration of the box are thus ẋe2 and ẍe2 =
the equation of motion for θ, the angle of the pendulum inside the box. To do so, we
will use Eq. (3.57), the relative-motion form of Newton’s second law.
The position of pendulum bob P in moving frame B = (O , b1, b2, b3) is
rP/O = l cos θb1 + l sin θb2.
86
CHAPTER THREE
Following the approach of Example 3.9, we find the velocity and acceleration of
the pendulum bob in the frame fixed to the box:
B
vP/O =
B
aP/O =
Bd
dt
Bd
dt
rP/O = −l θ̇ sin θe1 + l θ̇ cos θe2
I
vP/O = (−l θ̈ sin θ − l θ̇ 2 cos θ)e1 + (l θ̈ cos θ − l θ̇ 2 sin θ)e2.
Now use Eq. (3.57) and the free-body diagram in Figure 3.27c to obtain a modification of the two scalar equations in Example 3.9,
−F sin θ − mP a = mP (l θ̈ cos θ − l θ̇ 2 sin θ)
−F cos θ + mP g = mP (−l θ̈ sin θ − l θ̇ 2 cos θ).
By carefully considering the acceleration of the box in inertial space, we have
an additional term in the equations of motion. If the velocity of the box were constant (a = 0), then these equations would be identical to Eqs. (3.21) and (3.22) in
Example 3.9 because a frame moving at constant velocity is an inertial frame.
These equations can be manipulated as before to find the equation of motion
for θ:
θ̈ = −
g
a
sin θ − cos θ.
l
l
(3.58)
The equation of motion is more complicated because of the acceleration of the box.
There are two situations of interest. First, like the simple pendulum in Example 3.9,
we seek the equilibrium points for constant θ. To find the equilibrium angle θ0, set
θ̈ = 0 in Eq. (3.58) and solve for θ0:
a
.
θ0 = − arctan
g
Thus for a box accelerating at a, the pendulum hangs to the rear by an angle θ0,
as shown in Figure 3.28. This example is a model of a pendulous accelerometer—a
device used to measure acceleration. (It also describes well what happens to something
hanging from your car’s rear-view mirror when you accelerate.)
O′
g
l θ0
a
P
Figure 3.28 Equilibrium angle of a hanging pendulum in an accelerating box.
PLANAR KINEMATICS AND KINETICS OF A PARTICLE
87
We can also examine the motion for small θ, as in Example 3.9. Using the smallangle approximation, the linear equation for the accelerating pendulum is
θ̈ +
g
a
θ =− .
l
l
(3.59)
Eq. (3.59) is the equation for a simple harmonic oscillator with constant driving force
(discussed at the end of Tutorial 2.3). The solution is a sinusoidal motion about the
offset angle θ0.
3.7 How to Solve a Dynamics Problem
There are often many ways to solve a given dynamics problem. We recommend
starting with a general, rigorous approach to solve most problems, no matter how
hard (or easy) they appear. This approach, which is outlined below, utilizes tools from
our growing toolkit—vectors, reference frames, vector derivatives, and Newton’s
laws. Although we may not specifically refer to the following steps in examples and
tutorials, we seek to adhere to this process throughout the book.
1. Model the system. All dynamics problems start with a physical system to be
described. The first step is to produce a model of the system that incorporates
elements that can be handled mathematically, that is, point masses and/or rigid
bodies. Include any springs, strings, (massless) rods, hinges, and surfaces
that are relevant. There is no such thing as an “exact” model—as engineers
we are always making approximations. The key is to find the right set of
approximations that render the problem tractable but not trivial. Often this
step is the most important step. You should produce a drawing of the model.
2. Choose the coordinate(s). There should be at least one coordinate for each
degree of freedom of the system. Add the coordinates to your sketch of the
model or make a new sketch of the coordinates. For all coordinates, be sure
to indicate (e.g., with an arrow) the direction in which the coordinate is
increasing.
3. Define the reference frame(s). Define the inertial frame and any additional
frames that may be useful for solving the problem. Note that sometimes you
have to repeat this step! (The frames you initially choose may not turn out to
be the most useful.) For each frame, identify its origin and unit vectors. For
each rotating frame, write down its angular velocity with respect to the inertial
frame. If applicable, write down transformation tables among the unit vectors
in the various reference frames.
4. Draw the free-body diagram(s). For each particle and/or rigid body, draw
the free-body diagram and identify all forces and their lines of action. Label
the forces as vector components in your reference frames.
5. Calculate the kinematics. Write down the position of each particle (and the
position of the center of mass of each rigid body). Take the vector derivative
with respect to the inertial frame of the position to obtain the inertial velocity.
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CHAPTER THREE
Take the vector derivative with respect to the inertial frame of the velocity to
obtain the inertial acceleration.
6. Write down Newton’s second law. For each particle, write down Newton’s
second law in vector form using the free-body diagram and inertial acceleration. Using the transformation tables (Section 3.4.1) to convert the vector
components into a common frame, rewrite Newton’s second law as a system
of scalar algebraic equations.
7. Find the equations of motion. Compare the number of unknown variables—
forces and second-order time derivatives—to the number of equations. If
necessary, write down the constraint equation(s) and differentiate with respect
to time so that the number of scalar equations equals the number of unknowns.
Solve the system of algebraic equations for the unknown variables and write
down the equations of motion of the system.
8. Integrate the equations of motion. Integrate—analytically if you can, otherwise numerically—the equations of motion to find the trajectories of the
system.
This list may seem to have a lot of steps, but don’t despair. Once you start solving
problems, these steps will become automatic. The list is meant as a general guideline
and may not be applicable to every problem. However, it is important to always use
rigorous (vector) notation!
3.8 Derivations—Properties of the Vector Derivative
In this section we derive the various rules of vector derivatives shown in Table 3.1
using Definition 3.3. We also use them to derive the vector Taylor series.
3.8.1 The Addition Rule
The addition rule of scalar differentiation states that the derivative of the sum of two
functions is equal to the sum of their derivatives. It is easy to show that this rule also
holds for vector derivatives. Consider the vector derivative in a reference frame A of
the vector sum r(t) + s(t),
Ad
dt
(r(t) + s(t)) =
=
Ad
dt
Ad
dt
r1a1 + r2a2 + r3a3 + s1a1 + s2a2 + s3a3
(r1 + s1)a1 + (r2 + s2)a2 + (r3 + s3)a3 ,
where we have used the properties of vector addition in Appendix B.
Applying Definition 3.3 yields
Ad
dt
(r(t) + s(t)) =
d
d
d
(r1 + s1)a1 + (r2 + s2)a2 + (r3 + s3)a3
dt
dt
dt
= (ṙ1 + ṡ1)a1 + (ṙ2 + s˙2)a2 + (ṙ3 + ṡ3)a3,
PLANAR KINEMATICS AND KINETICS OF A PARTICLE
89
where we have used the distributive property of the scalar derivative. Finally, we
can again use the distributive property of vector addition from Appendix B and
Definition 3.3 to find
Ad
dt
(r(t) + s(t)) =
Ad
dt
r(t) +
Ad
dt
s(t),
(3.60)
which is the addition rule.
3.8.2 The Product Rule
Recall from calculus the scalar product rule:
d
(f (t)g(t)) = f˙(t)g(t) + f (t)ġ(t).
dt
For vectors, there are three possible products: the product of a vector and a scalar
function of time, the scalar dot product of two vectors, and the vector cross product
of two vectors.
Using the vector properties in Appendix B and the definition of the vector derivative, the derivative of a scalar times a vector is
Ad
dt
(f (t)r(t)) =
Ad
f (t)r1(t)a1 + f (t)r2(t)a2 + f (t)r3(t)a3
dt
d
d
d
= (f (t)r1(t))a1 + (f (t)r2(t))a2 + (f (t)r3(t))a3.
dt
dt
dt
From the scalar product rule we thus have
Ad
dt
(f (t)r(t)) = (f˙(t)r1(t) + f (t)ṙ1(t))a1 + (f˙(t)r2(t)
+ f (t)ṙ2(t))a2 + (f˙(t)r3(t) + f (t)ṙ3(t))a3
= f˙(t)(r1(t)a1 + r2(t)a2 + r3(t)a3) + f (t)(ṙ1(t)a1
+ ṙ2(t)a2 + ṙ3(t)a3),
where the distributive property of vector addition has again been applied.
Finally, applying Definition 3.3 gives
Ad
dt
(f (t)r(t)) =
Ad
df
r(t) + f (t) r(t),
dt
dt
which is the first product rule.
The derivative of the dot product of two vectors is just a scalar derivative,
d
d
r1(t)s1(t) + r2(t)s2(t) + r3(t)s3(t) ,
(r(t) . s(t)) =
dt
dt
(3.61)
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CHAPTER THREE
where we have used the definition of the dot product and expressed each vector as
components in the same frame. Using the distributive property of scalar differentiation
and the scalar product rule gives
d
(r(t) . s(t)) = ṙ1(t)s1(t) + r1(t)ṡ1(t) + ṙ2(t)s2(t) + r2(t)ṡ2(t)
dt
+ ṙ3(t)s3(t) + r3(t)ṡ3(t).
Rearranging leaves
d
(r(t) . s(t)) = ṙ1(t)s1(t) + ṙ2(t)s2(t) + ṙ3(t)s3(t) + r1(t)ṡ1(t)
dt
+ r2(t)ṡ2(t) + r3(t)ṡ3(t),
which, from the definition of the vector derivative and the dot product, gives the
second product rule:
Ad
Ad
d
(r(t) . s(t)) =
r(t) . s(t) + r(t) .
s(t).
dt
dt
dt
(3.62)
Of course, because the derivative of a dot product is a scalar derivative, the frame in
which we choose to take the vector derivatives is arbitrary—it just needs to be the
same for the two vectors.
The derivation of the product rule for the vector cross product follows a similar approach; we leave it to the problems (Problem 3.11). For completeness we state it here:
Ad
dt
(r(t) × s(t)) =
Ad
dt
r(t) × s(t) + r(t) ×
Ad
dt
s(t).
(3.63)
3.8.3 The Chain Rule
Now we consider the vector chain rule. Recall that, for a scalar function g(f (t)) of
another scalar function f (t), the derivative is
d
dg df
g(f (t)) =
.
dt
df dt
We consider three cases: a vector function of a scalar function of time, a scalar
function of a vector function of time, and a scalar function of a vector function of
time that has an explicit time dependence. The chain rule for the first case is a simple
generalization of the scalar chain rule,
Ad
dt
r(f (t)) =
Ad
df
r(f (t))
df
.
dt
The derivation again relies on Definition 3.3. We have
Ad
dt
r(f (t)) =
d d d r3(f (t)) a3.
r2(f (t)) a2 +
r1(f (t)) a1 +
dt
dt
dt
(3.64)
PLANAR KINEMATICS AND KINETICS OF A PARTICLE
91
Using the scalar chain rule on each component yields
Ad
dt
dr df
dr1 df
dr df
a1 + 2
a2 + 3
a3
df dt
df dt
df dt
dr
dr1
dr
= f˙(t)
a1 + 2 a2 + 3 a3 ,
df
df
df
r(f (t)) =
where we have again used the vector properties in Appendix B. The quantity in large
parentheses is the definition of the vector derivative with respect to f , which results
in Eq. (3.64).
The chain rule for a scalar function of a vector function of time is more subtle. To
begin, we have to be clear about what is meant by the notation f (r(t)). Since f is a
scalar function, to perform computations, its argument must be a set of scalars. Thus
f is operating on the scalar components of r(t) in some frame. Let r(t) in frame A be
r(t) = r1(t)a1 + r2(t)a2 + r3(t)a3.
Then the function f is written as
f (r1(t)a1 + r2(t)a2 + r3(t)a3) = f (r1(t), r2(t), r3(t)).
(3.65)
Now, even though f is a scalar function, its derivative is frame dependent, because
how it is written as a function of the scalar component magnitudes may differ depending on the frame we choose. In other words, taking the derivative of f with respect
to t is equivalent to asking how f depends on changes to r(t). But how r(t) varies
depends on which frame you choose, as seen throughout this chapter. Thus the definition of the scalar derivative of f (r(t)) with respect to time is similar to that of the
vector derivative. Taking the derivative with respect to frame A means we write r in
terms of components in A and f as a function of those components (as in Eq. (3.65))
and then use the scalar chain rule on each of the component magnitudes:
Ad
dt
f (r(t)) =
Ad
f (r1(t)a1 + r2(t)a2 + r3(t)a3) =
∂f
∂f
∂f
ṙ3
ṙ1 +
ṙ2 +
∂r3
∂r1
∂r2
dt
∂f
∂f
∂f
a3 . ṙ1a1 + ṙ2a2 + ṙ3a3 .
a1 +
a2 +
=
∂r3
∂r1
∂r2
Ad
=∇f
= dt r(t)
This calculation yields the following compact expression in terms of the gradient ∇f :
Ad
r(t).
f (r(t)) = ∇f .
dt
dt
Ad
(3.66)
A
Eq. (3.66) is the derivative of f in the direction of the vector dtd r. (See the discussion
of the directional derivative in Appendix A.) Note that, when taking the dot product,
the frame in which one expresses the components of each vector doesn’t matter.
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CHAPTER THREE
For a scalar function with an explicit time dependence (i.e., f (r(t), t)), the chain
rule becomes
Ad
Ad
∂f
f (r(t), t) = ∇f .
r(t) +
.
dt
dt
∂t
As a consequence, for any scalar function f (t) that depends only on time, the time
derivative is frame independent:
Ad
dt
f (t) =
Bd
dt
f (t) =
∂f
.
∂t
It is the latter expression that we are implicitly using when differentiating a coordinate
such as x(t) or θ(t).
Example 3.15 Calculating a Derivative Using the Vector Chain Rule
The chain rule derivation is subtle enough that it is helpful to see a simple example of
its use. Consider a point P in the plane with Cartesian coordinates (x, y)I and polar
coordinates (r, θ )I . Suppose the scalar function f is equal to the x-coordinate of P ,
f (rP/O ) = x = r cos θ,
(3.67)
where we have written f in terms of both Cartesian and polar coordinates. We now
compute the time derivative of f . It is, in this case, easy to just take the derivative
of the function in Eq. (3.67) to obtain f˙ = ẋ = ṙ cos θ − r θ̇ sin θ. However, let us
instead find it using the chain rule, Eq. (3.66). First find the gradient of f (rP/O ) (see
Appendix A):
∂f
∂f
∇f =
ey = e x
ex +
∂y
∂x
=
1 ∂f
∂f
er +
eθ = cos θer − sin θeθ .
∂r
r ∂θ
The time derivative of f in I , from Eq. (3.66), is
Id
Id
f (rP/O (t)) = ∇f .
rP/O
dt
dt
= ex . (ẋex + ẏey ) = ẋ
= (cos θer − sin θeθ ) . (ṙer + r θ̇ eθ ) = ṙ cos θ − r θ̇ sin θ = ẋ
= ex . (ṙer + r θ̇ eθ ) = ṙ cos θ − r θ̇ sin θ = ẋ.
Note that, as pointed out, the frame chosen for the components of the gradient or the
vector derivative doesn’t matter, as long as we are careful with the dot product.
Although the results agree with intuition, we are not quite done. What is the time
derivative of f (rP/O (t)) in the polar frame B? Using Eq. (3.66), we have
Bd
rP/O
f (rP/O (t)) = ∇f .
dt
dt
Bd
= (cos θer − sin θeθ ) . (ṙer )
= ṙ cos θ.
PLANAR KINEMATICS AND KINETICS OF A PARTICLE
93
This is a different result! The polar frame is defined such that P only changes its
position in one coordinate direction. Thus, if we are asking how f changes as rP/O
changes in B, the answer is that the change in f depends only on how r—but not
θ—changes.
3.8.4 The Vector Taylor Series
We wish to find a Taylor series expansion (Appendix A.4) of the vector function r(t)
about the point t = a. To do so, use the scalar Taylor series on each component. First,
write the vector r(t + a) as components in a frame A = (O, a1, a2, a3):
r(t + a) = r1(t + a)a1 + r2(t + a)a2 + r3(t + a)a3.
(3.68)
Then take the Taylor series of the scalar magnitude of each component. For i = 1, 2, 3,
we have
ri (t + a) = ri (a) +
dri
dt
(t − a) +
t=a
1 d 2 ri
2! dt 2
(t − a)2 + . . . .
t=a
Expanding and rearranging each coefficient in Eq. (3.68) yields
r(t + a) = r1(a)a1 + r2(a)a2 + r3(a)a3
+
+
dr1
dt
(t − a)a1 +
t=a
1 d 2 r1
2! dt 2
dr2
dt
(t − a)2a1 +
t=a
(t − a)a2 +
t=a
1 d 2 r2
2! dt 2
dr3
dt
(t − a)a3
t=a
(t − a)2a2 +
t=a
1 d 2 r3
2! dt 2
(t − a)2a3
t=a
+....
Using Definition 3.3, we obtain the vector Taylor series:
r(t) = r(a) +
+
Ad
dt
(t − a) +
r
1 Ad n
r
n! dt n
t=a
1 Ad 2
r
2! dt 2
(t − a)2 + . . .
t=a
(t − a)n + . . . .
t=a
3.9 Tutorials
Tutorial 3.1 A Torsion Spring
This tutorial examines the torsion spring—a mechanism that provides a restoring
force in a rotational direction. The torsion spring is an important modeling construct
analogous to a linear spring. Consider the configuration illustrated in Figure 3.29a.
Particle P is attached to a massless rod of fixed length l. The mass is acted on (by
means of the rod) by a torsion spring with restoring force −k(θ − θ0), where θ0 is
94
CHAPTER THREE
P
P
B
l
e2
–k(θ – θ 0)eθ
er
eθ
θ
O
I
P
O
k
(a) Coordinates
–Ter
e1
(b) Reference frames
(c) Free-body diagram
Figure 3.29 A torsion spring.
the rest angle of the spring.6 Using Figure 3.29c and the polar frame B illustrated in
Figure 3.29b, the total force on P is
FP = −T er − k(θ − θ0)eθ ,
where T is the tension in the rod. The kinematics of particle P are
rP/O = ler
I
vP/O = l θ̇eθ
I
aP/O = −l θ̇ 2er + l θ̈eθ .
Using Newton’s second law, we obtain
−T er − k(θ − θ0)eθ = mP (−l θ̇ 2er + l θ̈eθ ).
The tension in the rod,
T = mP l θ̇ 2,
acts as a centripetal force (see Section 3.5). The equation of motion,
θ̈ = −
k
(θ − θ0),
mP l
exhibits simple harmonic motion with natural frequency ω0 = mk l (see Tutorial 2.3
P
for the solution). As you might expect, this rotational mass-spring system oscillates
about the rest angle of the spring.
Tutorial 3.2 A Particle Sliding on a Parabola
This tutorial studies the dynamics of a particle sliding on a parabola, as shown in
Figure 3.30. This could be a model for a snowboarder in a half-pipe, as shown
6 Note
that k here has different units than the spring constant of the linear spring studied in Tutorial 2.3.
PLANAR KINEMATICS AND KINETICS OF A PARTICLE
I
I
g
ey
en
x
B
y = μx2
95
Nen
et
P
–mPgey
O
O
(a) Coordinates
ex
(b) Reference frames
(c) Free-body diagram
Figure 3.30 A particle sliding on a parabola.
Figure 3.31 A simple model for a snowboarder in a half-pipe is a particle sliding on a parabola,
the dynamics of which are explored in Tutorial 3.2. Image courtesy of Shutterstock.
in Figure 3.31. Define the (fixed) inertial frame I = (O, ex , ey , ez ) and the path frame
B = (P , et , en, ez ). Using the equation y(x) = μx 2 to describe the parabola and the
Cartesian coordinates shown in Figure 3.30a, the kinematics of P are
rP/O = xex + μx 2ey
I
vP/O = ẋex + 2μx ẋey
I
aP/O = ẍex + 2μ(ẋ 2 + x ẍ)ey .
Note that this system has one degree of freedom, x, since the y-position of the particle
is a function of x (y = μx 2 is a constraint equation).
Using the free-body diagram in Figure 3.30c, we write Newton’s second law for
the mass mP as
Nen − mP gey = mP ẍex + 2mP μ(ẋ 2 + x ẍ)ey .
(3.69)
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CHAPTER THREE
To separate this vector equation into two scalar equations and solve for the equation
of motion in ẍ, we must write en in terms of ex and ey (or vice versa). By definition,
we have
et =
Iv
P/O
Iv
P/O
ẋex + 2μx ẋey
ex + 2μxey
=
=
ẋ 2 + 4μ2x 2ẋ 2
1 + 4μ2x 2
−2μxex + ey
en = e z × e t = .
1 + 4μ2x 2
(3.70)
Substituting Eq. (3.70) into Eq. (3.69), we obtain the two scalar equations
−2μxN
= mP ẍ
1 + 4μ2x 2
N
1 + 4μ2x 2
− mP g = 2mP μ(ẋ 2 + x ẍ),
which yield the normal force
N =−
mP ẍ
2μx
mP (g + 2μẋ 2)
1 + 4μ2x 2 = 1 + 4μ2x 2
and the equation of motion
ẍ = −
2μx(g + 2μẋ 2)
.
1 + 4μ2x 2
(3.71)
We integrate the equation of motion using the matlab command ODE45. matlab
can also be used to calculate the normal force as a function of the position of P .
Since ODE45 integrates differential equations in first-order form—and the equation of
motion in Eq. (3.71) is a second-order differential equation—we rewrite the equation
[x, ẋ]T :
of motion using the change of variables Z = [z1, z2]T =
ż1 = z2
ż2 = −
2μz1(g + 2μz22)
1 + 4μ2z12
.
(3.72)
In matlab, we create the function MYODEFUN to define the equations of motion
(Eq. (3.72)):
function zdot = myodefun(t,z,g,mu)
zdot(1,1) = z(2);
zdot(2,1) = -2*mu*z(1)*(g+2*mu*z(2)^2)/(1+4*mu^2*z(1)^2);
Next we define the constants g and mu in the Command Window and call ODE45 to
integrate the equations of motion from x(0) = 1 m and ẋ(0) = 0 m/s for 10 s:
>> [t,z] = ode45(@myodefun,[0 10],[1 0],[],g,mu);
PLANAR KINEMATICS AND KINETICS OF A PARTICLE
5
x (m)
5
x (m/s)
0
–5
0
97
x (m)
x (m/s)
0
2
4
6
8
–5
0
10
2
t (s)
4
6
8
10
t (s)
N(N)
50
N(N)
50
0
–1.0
– 0.5
0.0
x (m)
0.5
1.0
0
–1.0
– 0.5
(a)
0.0
x (m)
0.5
1.0
(b)
Figure 3.32 Particle sliding on a parabola. The top panels show position x(t) and speed
ẋ(t); the bottom panels show normal force N as a function of position x. (a) μ = 0.1 m−2 .
(b) μ = 1.0 m−2 .
The first column of the output matrix z contains the position x(t) and the second
column of z contains the speed ẋ(t). To compute the normal force N , enter
>> N = m*(g+2*mu*z(:,2).^2)./sqrt(1+4*mu^2*z(:,1).^2);
Note the use of the . for pointwise matrix operations.
The results are plotted in Figure 3.32 for x(0) = 1 m, ẋ(0) = 0 m/s, mP = 1 kg,
and two values of the parabola parameter—μ = 0.1 m−2 and μ = 1.0 m−2. For the
initial position x(0) = 1 m, the initial height satisfies y(0) = μx(0)2 = μ m. For
μ = 0.1 m−2, both the position and the speed oscillate sinusoidally, and the normal
force only slightly exceeds the weight mP g of the particle. For μ = 1m−2, the position
still oscillates, but the speed curve has sharper peaks than a sinusoid; the normal force
exceeds five times the weight of the mass.
Tutorial 3.3 A Particle Sliding on a Hemisphere
This tutorial examines the dynamics of a particle sliding on a frictionless hemisphere
of radius R, as shown in Figure 3.33. We solve analytically for the angular displacement of the particle at the instant it falls off the hemisphere and use matlab to compute
the time at which the particle falls off.
We use the (fixed) inertial frame I = (O, ex , ey , ez ) and the rotating polar frame
B = (O, er , eθ , e3), where e3 = −ez . The unit vectors in frames I and B are related
by the coordinate transformation table
ex
er
eθ
sin θ
cos θ
ey
cos θ
− sin θ .
The particle starts at θ = 0 and is given a very small push in the direction of increasing
θ. At some time t = t ∗ later, the particle will lose contact with the surface at angular
98
CHAPTER THREE
I
I
ey
g
θ
B
Ner
P
–mPgey
er
r=R
ex
O
O
(a) Coordinates
eθ
(b) Reference frames
(c) Free-body diagram
Figure 3.33 A particle sliding on a hemisphere.
displacement θ ∗ = θ (t ∗). Before it falls off the surface, the kinematics of particle
P are
rP/O = Rer
I
vP/O = R θ̇eθ
I
aP/O = R θ̈eθ − R θ̇ 2er ,
where we used Ṙ = R̈ = 0.
As usual, let mP denote the mass of particle P . Using the free-body diagram in
Figure 3.33c and Newton’s second law yields
Ner − mP gey = mP (R θ̈eθ − R θ̇ 2er ).
Using the coordinate transformation table to write ey in terms of er and eθ , we can
solve for the normal force N and the equation of motion of the system:
N = mP g cos θ − mP R θ̇ 2
g
θ̈ = sin θ.
R
(3.73)
(3.74)
Note that the equation of motion √
Eq. (3.74) looks like the equation for simple harmonic
motion with natural frequency g/R, except that the sign of the right-hand side is
reversed—if it was physically realizable, the particle would oscillate about θ = π
instead of θ = 0.
Perhaps counterintuitively, the particle falls off the surface at the instant the normal
force satisfies N = 0 (or, equivalently, N/mP = 0). At this instant, the angular rate
θ̇ = θ̇ ∗ satisfies the equation
0 = mP g cos θ ∗ − mP R(θ̇ ∗)2,
that is,
∗
θ̇ =
g
cos θ ∗.
R
(3.75)
We solve analytically for the angle θ ∗ by integrating Eq. (3.74) from 0 to t. That is,
using separation of variables (see Section C.2.1) and the initial conditions θ(0) = 0
PLANAR KINEMATICS AND KINETICS OF A PARTICLE
99
and θ̇ (0) = θ̇0 > 0 gives
1 2 1 2
g
θ̇ − θ̇0 = − (cos θ − 1),
2
2
R
which implies
θ̇02 +
θ̇ =
2g
(1 − cos θ).
R
(3.76)
Evaluating Eq. (3.76) at t ∗ and equating it to Eq. (3.75), we obtain the relation
cos θ ∗ =
2
R
+ θ̇ 2,
3 3g 0
which implies
∗
θ = arccos
2
R
+ θ̇02 .
3 3g
(3.77)
In the limit θ̇0 → 0 (i.e., if the initial push is very small), θ ∗ ≈ arccos(2/3).
Next solve for the time t ∗ by numerically integrating Eq. (3.76) assuming R =
1 m and θ̇0 = 0.01 m/s (note that we cannot integrate Eq. (3.75), which holds only at
the instant t = t ∗). Using separation of variables, we obtain
t∗ =
θ∗
0
dθ
θ̇02
+
,
(3.78)
2g
R (1 − cos θ)
where θ ∗ is given by Eq. (3.77). To integrate the right-hand side, use the matlab command QUAD, which is short for quadrature.7 Just like ODE45, which integrates equations of motion contained in a separate function, the QUAD function uses an integrand
contained in a separate function QUADFUN. The QUAD function is invoked by passing the function handle @QUADFUN using the syntax Q = QUAD(@(X)QUADFUN(X,P1,
P2, . . . ),A,B), where X is the argument of the integrand function FUN, P1 and P2
are parameters to FUN, and A and B are the integral limits. In our case, FUN contains
the integrand in Eq. (3.78):
function dt = myquadfun(theta,g,R,dottheta0)
dt = 1./sqrt(dottheta0^2+2*g/R*(1-cos(theta)));
Note that we need to use the array divide ./ in this function because theta will
be an array of values from 0 to θ ∗. We define the following variables in
the matlab workspace: g=9.81 m/s2, R=1 m, dottheta0=0.01 rad/s, and
thetastar=acos(2/3+R/3/g*dottheta0^2) rad. Finally, call QUAD from the command line
>> quad(@(theta)myquadfun(theta,g,R,dottheta0),0,thetastar),
which returns t ∗ = 2.0 s.
7 “Quadrature”
is a term that can mean solving an integral analytically or numerically.
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CHAPTER THREE
3.10 Key Ideas
.
.
A vector indicates a magnitude and a direction in space. A unit vector is a vector
with magnitude one. Three orthogonal unit vectors—along with an origin—define
a reference frame, for example, A = (O, a1, a2, a3).
A vector can be written as the sum of weighted unit vectors—called vector
components. For example, the position of point P with respect to O is
rP/O = a1a1 + a2a2 + a3a3.
A vector can be written in matrix notation using the magnitude of the vector
components:
⎡ ⎤
a1
⎣
[rP/O ]A = a2 ⎦ .
a3 A
.
If frame B = (O, b1, b2, b3) is rotated about the b3 axis by an angle θ with respect
to frame A, then the unit vectors of A and B are related by the transformation
table
a1
a2
b1
b2
.
.
cos θ
− sin θ
sin θ
cos θ .
A coordinate system is a set of scalars that specifies the position of a point in a reference frame. Three examples of a coordinate system are Cartesian coordinates,
polar coordinates, and path coordinates.
The rate of change with time of a vector is reference-frame dependent. The velocity
Ad
Av
P/O = dt (rP/O ) is the rate of change of the position rP/O in reference frame A.
A
d A
Likewise, the acceleration AaP/O =
dt ( vP/O ) is the rate of change of the velocity
in reference frame A. Since the unit vectors a1, a2, a3 are fixed in frame A, the
velocity (or acceleration) of P with respect to frame A is given by differentiating
the scalar weights of each component:
A
vP/O = ȧ1a1 + ȧ2a2 + ȧ3a3.
.
The rate of rotation of the polar frame (or path frame) B with respect to frame A
is given by the simple angular velocity AωB . Since B rotates with respect to A
about a common axis b3 = a3,
A B
θ̇a3,
ω = AωB a3 =
where AωB =
θ̇ is the angular speed.
.
The rate of a change in frame A of unit vector bi , i = 1, 2, that is fixed in frame
B is
Ad
dt
(bi ) = AωB × bi ,
PLANAR KINEMATICS AND KINETICS OF A PARTICLE
101
where the angular velocity AωB corresponds to rotation of B about a common axis
b3 = a3.
.
Two important frames are the polar frame B = (O, er , eθ , e3) and the path frame
B = (P , et , en, e3). The inertial velocity and inertial acceleration expressed in the
polar frame using polar coordinates (r, θ )I are
I
vP/O = ṙer + r θ̇eθ
I
aP/O = (r̈ − r θ̇ 2)er + (2ṙ θ̇ + r θ̈)eθ .
.
The inertial velocity and inertial acceleration expressed in the path frame using
path coordinate s are
I
vP/O = ṡet
I
aP/O = s̈et +
ṡ 2
en ,
Rc
where Rc is the instantaneous radius of curvature of the path.
.
The trajectory of a particle, P , can be found relative to frame B with origin O even
if O is accelerating in the inertial frame by using the modified form of Newton’s
second law
mP BaP/O = FP − mP IaO /O .
.
The vector derivative obeys many of the same properties as the scalar derivative,
including the addition rule, product rule, and chain rule.
3.11 Notes and Further Reading
Nearly all introductory (and advanced) dynamics texts treat kinematics of particles,
but there is great variation in the notation and level of detail. Very few emphasize
reference frames as we do here; those that do (e.g., Kane and Levinson 1985; Greenwood 1988; Tenenbaum 2004) are typically used as graduate texts. Rao (2006) has a
particularly nice discussion of the importance and definition of reference frames and
was a reference for our mathematical and operational definitions. As noted in Chapter 1, our notation is similar to that developed by Kane (1978) and Kane and Levinson
(1985) and used by Rao (2006) and Tenenbaum (2004).
Our use of the unit-vector transformation array is modeled after the approach
taken in Kane (1978), Kane and Levinson (1985), and Tongue and Sheppard (2005).
Tutorial 3.2 is based on a similar example in Rao (2006). Many texts take a geometric
approach to the definition of the vector derivative. We follow the analytical approach
in Kane (1978).
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CHAPTER THREE
3.12 Problems
3.1
Express rP/Q shown in Figure 3.34 using vector components in frame A. Now
express rP/Q using vector components in frame B.
P
B
A
a2
4
b2
Q
3
b1
30°
O
a1
Figure 3.34 Problem 3.1.
3.2
Three forces act on particle P : F1 = 2ex − 5ey N, F2 = 10(cos θex + sin θey )
N, and F3 = −20ey N. If the resultant force is zero, find θ and the force vector
F2. Sketch the three forces acting on P .
3.3
Consider frames I = (O, e1, e2, e3) and A = (O, a1, a2, a3), where e3 = a3,
as shown in Figure 3.35. Find the position of P with respect to O in the
following coordinates:
a.
b.
c.
d.
Cartesian coordinates in I , (x, y)I .
Cartesian coordinates in A, (a1, a2)A .
Polar coordinates in I , (r, θ)I .
Polar coordinates in A, (ρ, β)A .
I
A
e2
2.3
P
a2
a1
3.1
30°
O
e1
Figure 3.35 Problem 3.3.
3.4
In Example 3.5, find expressions for the polar coordinates (r, θ )I in terms of
the path coordinate s.
3.5
The position of P shown in Figure 3.36 is described by the path coordinate s
on a circular path with radius ρ and whose center Q is located at rQ/O = ae1.
Find the position of P with respect to O using Cartesian coordinates in frame
I , (x, y)I .
PLANAR KINEMATICS AND KINETICS OF A PARTICLE
103
I
e2
P
ρ
a
O
Q
s
e1
Figure 3.36 Problem 3.5.
3.6
Consider a two-link arm shown in Figure 3.37 with frames I = (O, e1, e2, e3),
A = (O, a1, a2, a3), and B = (O, b1, b2, b3), where e3 = a3 = b3. Construct
three transformation tables, relating the unit vectors of frames I and A, A
and B, and I and B.
B
β
A
I
α
Figure 3.37 Problem 3.6.
3.7
Sketch an idealized planar model of the crane shown in Figure 3.38 using a
point mass, a rigid, massless rod, and a massless cable. How many degrees of
Figure 3.38 Problem 3.7. Image courtesy of Shutterstock.
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CHAPTER THREE
freedom are there in your model? Derive the inertial kinematics of a point on
the hook of the crane.
3.8
Consider inertial frame A = (O, a1, a2, a3) and polar frame B = (O, b1, b2,
b3), as shown in Figure 3.39, where a3 = b3 and B is rotating with respect
to A with angular velocity AωB = θ̇b3. The position of P with respect to O
expressed as vector components in frame B is rP/O (t) = b1(t)b1. Find BvP/O
and AvP/O .
A
B
a2
b2
P
b1
θ
O
a1
Figure 3.39 Problem 3.8.
3.9
Consider the example discussed in the beginning of Section 3.5 and shown
in Figure 3.23. Suppose the constant inertial velocity is given in Cartesian
coordinates as:
I
vP/O = v1e1 + v2e2
a. Calculate the rates of change ṙ and θ̇ of the polar coordinates in terms
of v1, v2, r, and θ.
b. Using Eq. (3.51), find the second derivatives r̈ and θ̈ of the polar
coordinates in terms of v1, v2, r, and θ.
c. Show that asymptotically (as time goes to infinity) θ̇ and θ̈ go to zero
and the polar frame reaches a fixed orientation in the inertial frame
with the radial unit vector er parallel to the velocity.
3.10 Use the angular velocity to derive the expression for the acceleration of a
particle in polar coordinates in Eq. (3.33).
3.11 Show that the product rule for vector differentiation (see Table 3.1) holds for
the vector cross product.
3.12 Derive the vector derivative of eθ in Eq. (3.30).
3.13 Show that the radius of curvature of a path defined by the function y(x) in
Cartesian coordinates is given by Eq. (3.50).
3.14 Consider the inertial frame I = (O, e1, e2, e3) and a point P that is constrained
to move in the (e2, e3) plane, as shown in Figure 3.40.
a. Sketch a polar frame B and derive the transformation table between B
and I .
b. Write the position rP/O of P with respect to O in polar coordinates
using vector components in frame B.
PLANAR KINEMATICS AND KINETICS OF A PARTICLE
105
I
e3
P
O
e2
e1
Figure 3.40 Problem 3.14.
c. What is the direction of angular velocity IωB of B with respect to I ?
What is its magnitude?
d. Derive the inertial kinematics of P with respect to O in polar coordinates using vector components in frame B.
3.15 Consider a bead P that is constrained to move along a path x = y 2, as shown
in Figure 3.41, where (x, y)I are Cartesian coordinates in frame I . Assuming
the bead is forced to move at constant rate v0 in the e2 direction, write IvP/O
and IaP/O in Cartesian coordinates as vector components in frame I .
I
e2
I
vP/O x = y 2
v0
P
O
e1
Figure 3.41 Problem 3.15.
3.16 Consider a particle P accelerating from rest on a curved path with s̈ = a,
where a is a constant.
a. Find the radius of curvature Rc (t) such that the magnitude of the
particle’s acceleration IaP/O is equal to 2g for all time.
b. Sketch the path of the particle.
3.17 Let A = (O, a1, a2, a3) be an inertial reference frame and B = (O, b1, b2, b3)
be a polar reference frame with AωB = θ̇b3, as shown in Figure 3.42. Derive
the following in polar coordinates (r, θ )A :
a. The transformation table between A and B.
b. The position rP/O of P with respect to O expressed as components
in B.
c. The position rP/O of P with respect to O expressed as components
in A.
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CHAPTER THREE
A
B
a2
b2
P
r
b1
θ
O
a1
Figure 3.42 Problem 3.17.
d. The velocity BvP/O and acceleration BaP/O of P with respect to O in
B expressed as components in B.
e. The velocity AvP/O and acceleration AaP/O of P with respect to O in
A expressed as components in B.
3.18 Let I = (O, e1, e2, e3) be an inertial reference frame and frame A =
(O , a1, a2, a3) be aligned with I and translating with velocity IvO /O = v0e1.
Frame B = (O , b1, b2, b3) is a polar frame with angular velocity IωB = θ̇b3.
(P is free to move in A.) Derive the following using the coordinates shown
in Figure 3.43:
a. The transformation table between A and B.
b. The position rP/O of P with respect to O expressed as components
in B.
c. The position rP/O of P with respect to O expressed as components
in B.
d. The velocity BvP/O and acceleration BaP/O of P with respect to O in
B expressed as components in B.
e. The velocity AvP/O and acceleration AaP/O of P with respect to O in
A expressed as components in B.
I
B
e2
A
b2
a2
P
r
x
b1
θ
O
e1
O′
v0
a1
Figure 3.43 Problem 3.18.
3.19 To simulate the zero-gravity conditions of space, the National Aeronautics
and Space Administration (NASA) regularly flies an airplane on a parabolic
trajectory. At the apex of the trajectory, gravity seems to disappear and everything in the cabin floats for about 30 s. Consider such a parabolic trajectory
PLANAR KINEMATICS AND KINETICS OF A PARTICLE
107
given by y = −cx 2. While the plane is flying up the parabola, you are stationary, that is, you feel a normal force N that cancels your weight. At the top of
the parabola, when you begin to float, the normal force vanishes and you (and
the plane) are in free fall as if you were in orbit.
a. Draw a free-body diagram of you in the plane.
b. Find your acceleration as a function of the speed v of the airplane,
which you may assume is constant, and the shape c of the parabola.
c. Find an expression for c in terms of v, so that zero-gravity conditions
are achieved at the apex of the trajectory.
3.20 Consider a collar of mass m sliding on a frictionless shaft, as depicted in
Figure 3.44. The collar can slide along the shaft and is connected to a spring
of spring constant k, the other end of which is connected to the pivot point O
a distance l from the shaft. The unstretched length of the spring is l.
a. How many degrees of freedom are there in this problem?
b. Find the equation of motion for the collar.
O
l
k
m
C
x
Figure 3.44 Problem 3.20: Sliding collar on a spring.
3.21 A snowboarder of mass mB and cross-sectional area A is sliding down the
terrain park, which has lots of powder. Draw a free-body diagram of the
boarder identifying all forces acting on her. If you were asked to figure out
whether she was going to leave the ground on a particular hill, what coordinate
system would you choose to solve the problem?
3.22 Using the equations of motion for the simple pendulum in Cartesian coordinates (Eq. (3.7)), numerically integrate the trajectory for the initial conditions
y(t)
and
(θ0, θ̇0) = (0.1 rad, 0 rad/s). Plot the pendulum angle θ(t) = arctan x(t)
the pendulum length l(t) = x(t)2 + y(t)2 as a function of time. What do
you notice about the length?
3.23 Consider the simple planar pendulum discussed at length in this chapter, only
now, rather than a solid massless rod, suppose the pendulum bob is attached
to a spring with spring constant k (Figure 3.45). Note that the spring does not
bend or twist.
a. How many degrees of freedom are there in this problem?
b. What coordinates would you use to describe the configuration of the
pendulum bob?
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CHAPTER THREE
Figure 3.45 Problem 3.23.
c. Solve for the equations of motion in the coordinates you chose in (b).
Be sure to draw a free-body diagram.
3.24 Solve the equations of motion for the spring-pendulum system in the previous
problem using matlab ODE45 and plot the pendulum angle θ(t) and pendulum
length r(t) versus time. Assume that the pendulum is released with initial
conditions θ (0) = π/4, θ̇(0) = 0, r(0) = l0, and ṙ(0) = 0, where l0 is the rest
length of the spring. Let m = 1 kg, l0 = 0.1 m, and k = 2 N/m. Be sure to label
your plots with units and include a legend.
3.25 Derive the inertial kinematics of the tip of a two-link arm, shown in Figure 3.37.
3.26 Consider the planar spring in Example 3.12. Let k/mP = 10 s−2 and r0 =
1 m. Computationally simulate a 5-s trajectory of the system for the initial
conditions r(0) = 5 m, ṙ(0) = 0, θ(0) = π/4 rad, and θ̇(0) = π/6 rad/s.
3.27 Rederive the equations of motion for the planar spring in Example 3.12
including a Coulomb friction force acting opposite the velocity.
3.28 Suppose a rocket with mass 104 kg propels itself at an angle of 30◦ above the
horizontal using a thrust force of 2 × 107 N, as shown in Figure 3.46. Use a
path frame to compute the magnitude of the tangential and normal components
of the rocket’s acceleration at the instant shown in the figure.
T
W
30°
Figure 3.46 Problem 3.28.
3.29 Sketch a planar model of a two-cord bungie jump (Figure 3.47) in which you
represent the person of mass mP using a point mass P and the bungie cords as
simple springs with identical spring constants k and rest lengths l0. Assume
the distance between the attachment points is 2l0.
PLANAR KINEMATICS AND KINETICS OF A PARTICLE
109
Figure 3.47 Problem 3.29. Image courtesy of Shutterstock.
a. How many degrees of freedom does your model have?
b. Find the equations of motion of P .
c. Integrate the equations of motion in matlab using mP = 50 kg, k =
10 N/m, and l0 = 5 m, assuming that the person falls from rest.
d. Plot the tension in the bungie cords versus time. How strong do they
have to be so they won’t snap?
3.30 Consider a lunar rover of mass m = 200 kg traveling at constant speed over
a semicircular hill of radius ρ = 100 m. The acceleration due to gravity on
the moon is g = 1.6 m/s2. How fast can the rover travel without leaving the
moon’s surface anywhere on the hill?
3.31 Consider a mass m attached to the end of a rope of length l. One end of the
rope is fixed. If the mass travels at a constant speed v0 in a circle centered on
the fixed end of the rope (with the rope taut), what is the magnitude of the
centripetal acceleration on the mass?
3.32 Consider a race car being transported in a moving trailer, as shown in Figure 3.48. Suppose the trailer is traveling at 30 mph when it hits a bump that
jostles the race car off of its blocks and opens the trailer ramp. The race car
starts to accelerate down the trailer ramp, which is 1 m tall and 5 m long. What
110
CHAPTER THREE
g
30 mph
5m
1m
Figure 3.48 Problem 3.32.
is the inertial velocity of the car when it hits the ground? In what direction does
it roll (left or right)? [HINT: Model the car as a point mass.]
3.33 Consider a mass-spring system in a falling box (Figure 3.49). Let mP = 1 kg,
the spring constant k = 2 N/m, and the rest length of the spring l0 = 0.2 m.
Find the equations of motion of the mass.
k, l0
g
P
mP
Figure 3.49 Problem 3.33.
3.34 Consider a helium balloon attached by a taut string to an inertially fixed point
(Figure 3.50). Assume that the buoyancy force is equal to twice the weight of
Figure 3.50 Problem 3.34.
PLANAR KINEMATICS AND KINETICS OF A PARTICLE
111
the balloon. Find the equations of motion of the balloon relative to the fixed
point using the following steps:
a.
b.
c.
d.
e.
Sketch a point-mass model of the balloon and string.
Choose coordinates and define reference frame(s).
Draw a free-body diagram of the balloon.
Derive the inertial kinematics of the balloon.
Write down Newton’s second law for the balloon and solve for its
equation(s) of motion.
3.35 Suppose you forget to tie down a load of mass m on a flatbed truck of mass M,
as in Figure 3.51. Suppose the truck starts to move forward with acceleration
a. What does the coefficient of static friction have to be for the load not to
slide off the back of the truck?
m
a
Figure 3.51 Problem 3.35.
3.36 Suppose a batter hits a fly ball that leaves his bat with speed v0 at angle θ0 with
the horizontal (Figure 3.52). Assuming that the ball is initially at a height of
h0 above the ground and there is no air resistance, how high, h, does it go?
How far, d, does it go?
v0
g
θ0
h
h0
d
Figure 3.52 Problem 3.36.
3.37 A mass m is on an inclined plane at angle θ, as shown in Figure 3.53. Attached
to the mass is a spring of spring constant k that is fixed at the top of the incline.
There is gravity in this problem.
a. Assuming a coefficient of friction μ, find the equation of motion for
this system using Cartesian coordinates in the A frame. [HINT: Use
the sgn(.) function to extract the sign of the speed.]
b. Solve the equation of motion assuming no friction with initial conditions x(0) = x0 and ẋ(0) = ẋ0.
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CHAPTER THREE
a2 A
O
a1
k
m
θ
Figure 3.53 Problem 3.37.
3.38 Consider the classic problem of boring a tunnel through the earth (say, from
Tokyo to New York). Figure 3.54 shows such a tunnel bored at colatitude
λ (the colatitude is the angle from the north). Assume a mass P can move
freely in the tunnel, there is no friction, and the earth is not rotating. The only
force acting on P is gravity. Show that the equation of motion for P is simple
harmonic motion. What is the frequency of the oscillation?
P
λ R
O
Figure 3.54 Problem 3.38.
[HINT: R is the radius of the earth (6,378.145 km). It is a well-known result
(that we haven’t discussed) that the gravitational attraction on a particle inside
the earth is due only to the mass inside the radial position of the particle. In
other words, there is no gravitational force inside a spherical shell. Assume
the earth has uniform density ρ, and ignore the effect of the missing material
from the tunnel (it is small).]
CHAPTER FOUR
Linear and Angular Momentum of a Particle
Chapter 3 provided all of the basic tools you need to find the equations of motion for
any (planar) particle dynamics problem starting from Newton’s second law. Nevertheless, there are many additional concepts that can make finding equations of motion
easier. Using these concepts, we can sometimes solve for the trajectory of a particle—
or properties of the trajectory—without having to find and integrate the equations of
motion. This chapter explores two of these additional concepts that lend new insights
into the motion and provide new tools for solving specific problems. Nevertheless,
the only physics is still Newton’s three laws.
4.1 Linear Momentum and Linear Impulse
Recall from Chapter 2 the linear momentum form of Newton’s second law
FP =
Ip
Id
dt
pP/O ,
I
Iv
where P/O = mP P/O is the linear momentum. Integrating this equation with
respect to time for an arbitrary force FP yields
t2
t2 I
FP dt =
d IpP/O = IpP/O (t2) − IpP/O (t1)
(4.1)
t1
t1
or, after rearranging and using the definition of linear momentum,
mP IvP/O (t2) = mP IvP/O (t1) +
t2
FP dt.
(4.2)
t1
Eqs. (4.1) and (4.2) state that the change in linear momentum of a particle P is equal
to the integrated force acting on P . Unfortunately, the integral on the right side of
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CHAPTER FOUR
Eq. (4.2) usually is too hard to solve or depends on the position or velocity (the state
of the particle), making Eq. (4.2) a differential equation. However, there are three
situations where this equation is useful:
a. FP is identically zero. That is, the total force acting on particle P is zero. In
this case, the magnitude and direction of IpP/O are constant over the trajectory
of P . This case is simply the law of conservation of linear momentum (or
Newton’s first law) discussed in Section 2.2.5.
b. FP is constant or is a known function of time.That is, FP = 0 is independent of
the position and velocity of the particle. In this case, the integral in Eq. (4.2)
can be solved, either explicitly or numerically, to relate the initial and final
velocities at t1 and t2, respectively. Constant, uniform gravity is an example
of such a force (e.g., Tutorial 2.6).
t2 − t1 is extremely short.That is, FP acts over a very small time interval.
c. t =
We call such a force an impulsive force. In this case—which is the subject of
the rest of this section—we can safely ignore the details of the dependence of
FP on the particle state to solve Eq. (4.2).
We begin our discussion of the third case with the definition of linear impulse.
Definition 4.1 The linear impulse FP (t1, t2) acting on particle P is the integrated
t2 − t1,
force on P over a (short) time interval t =
FP (t1, t2) =
t2
FP dt.
t1
Figure 4.1a shows a schematic of the magnitude FP of an impulsive force and
the area underneath its plot with respect to time—which is the magnitude FP (t1, t2)
of the impulse. Remember that the impulse is a vector and thus also has a direction.
Linear impulse is an extremely useful concept. Rather than try and understand
the details of what might be a very complex force, we simply ignore the motion
during application of the force and treat the problem as an instantaneous change in
the momentum, via Eq. (4.2), due to the linear impulse. In this model, the impulse
is so short that we assume the change in linear momentum of particle P occurs
without significant motion of P . This assumption is illustrated in Figure 4.1b and c.
In fact, this assumption was implicit in the solution to the tossed point-mass problem
in Tutorial 2.6. Recall that we neglected the detailed modeling of the force during
the toss and simply computed the trajectory as a function of the initial conditions.
The tossing force can be treated as an impulse whose effect is to change the initial
conditions. The goal of the thrower is to apply just the right impulse to achieve the
desired initial velocity. After the release of the particle, we can solve the equations of
motion for the trajectory under gravity alone.
Why is it important to assume the force acts only over a short interval? Isn’t
Eq. (4.2) exact? Yes, it is. However, most forces depend on the state of the particle. If
the time interval t is small, then we can treat the force as integrable and avoid solving
LINEAR AND ANGULAR MOMENTUM OF A PARTICLE
115
–
||
||F
P (t1,t2)
t2
t1 ≈ t2
v(t)
x(t)
||FP(t)||
t1
t1 ≈ t2
t
(a)
t
(b)
t1 ≈ t2
t
(c)
Figure 4.1 Linear impulse FP (t1, t2 ) is the integral from t = t1 to t = t2 of the force FP
that produces an instantaneous change of direction of motion and a discontinuous change
in velocity. (a) An impulsive force. (b) Instantaneous change of direction. (c) Discontinuous
change in velocity.
a differential equation. It is rare that a force is independent of position and velocity
over long periods of time. However, when the force on a particle is impulsive and acts
only over a very short interval, modeling it as state independent—and assuming the
particle position doesn’t change—is an excellent approximation.
Example 4.1 Impulsive Spring
One of the simplest examples of a linear impulse is the sudden striking of a mass on
a spring (see Tutorial 2.3 for the solution of the motion of a mass on a spring). This
example is an excellent model of, for example, the motion of a car tire hitting a bump
in the road. The car is massive enough that the tires can be considered connected to
a fixed wall by means of a spring (and, possibly, a shock absorber).
Consider the model illustrated in Figure 4.2. Suppose the mass is initially at rest.
It is struck by a force that produces a linear impulse of magnitude F P (t1, t2) in
the negative x-direction. This impulse has units of N-s. In our idealized model of
the impulse, the mass does not move until after time t2. Using Eq. (4.2), the mass
has horizontal speed ẋ(t2) = −F P (t1, t2)/mP after the impulse. Assuming that the
I
ey
x
–
FP(t1,t2)
–kxex
m
ex
Figure 4.2 A mass on a spring hit by impulse FP (t1, t2 ) in the −ex direction.
116
CHAPTER FOUR
I
I
vP/O
I
B
rQ/P
vQ/O
Q
P
O
(a)
(b)
Figure 4.3 Synchronized swimming. (a) Top view. (b) Reference frames.
position of the mass at t = t2 is x(t2) = 0, then we can simply substitute the speed
ẋ(t2) into the trajectory Eqs. (2.22) and (2.23) to find the final motion of the mass:
x(t) =
−F P (t1, t2)
sin(ω0t)
mP ω0
(4.3)
ẋ(t) =
−F P (t1, t2)
cos(ω0t).
mP
(4.4)
Thus the linear impulse F P (t1, t2) sends the mass-spring system into oscillatory
motion with amplitude proportional to the magnitude of the impulse.
Example 4.2 Synchronized Swimming
Consider the two fish shown in Figure 4.3a. Each fish glides through the water, beating
its tail intermittently. Suppose that, between tail beats, each fish glides at constant
velocity. The interaction between a flapping tail and the water is quite complicated;
nonetheless we can study the fish motion using the concept of linear impulse.
Model the fish using two point masses P and Q. Assume Q is gliding at a constant
velocity IvQ/O , and suppose fish P seeks to align with fish Q, knowing only the
velocity of Q relative to its velocity. That is, fish P cannot estimate the inertial velocity
Iv
B
Q/O , only its relative velocity vQ/P in a path coordinate frame fixed to P . Let’s
find the linear impulse that fish P needs to exert to match inertial velocities with fish
Q. Remember, by Newton’s third law, if a fish exerts an impulse on the water with
its tail, the water then exerts an equal and opposite linear impulse that accelerates the
fish.
From Figure 4.3b we have
rQ/O = rQ/P + rP/O ,
which implies
vQ/O = IvQ/P + IvP/O .
I
(4.5)
Between tail beats, the path frame B is inertial, which implies IvQ/P = BvQ/P . If fish
P beats its tail from time t1 to t2, it seeks to satisfy
LINEAR AND ANGULAR MOMENTUM OF A PARTICLE
117
I
vP/O (t2) = IvQ/O (t1)
= BvQ/P (t1) + IvP/O (t1).
Using Eq. (4.2) and Definition 4.1, the linear impulse required to instantaneously
change the velocity of P to align with Q is
FP (t1, t2) = mP IvP/O (t2) − IvP/O (t1)
= mP BvQ/P (t1).
(4.6)
Note the smaller the speed of Q relative to P , the smaller the magnitude of linear
impulse required to align them. (Of course a real fish is not a point mass and thus
would also want to rotate its body while changing its velocity direction.)
Eq. (4.6) also looks a bit like a linear momentum. To examine this, we observe that
vP/O = IvP/Q + IvQ/O ,
I
which, using Eq. (4.5), implies IvQ/P = −IvP/Q. Therefore an equivalent expression
for the linear impulse in Eq. (4.6) is
FP (t1, t2) = −mP BvP/Q(t1) = −BpP/Q(t1).
As we might have expected, the applied linear impulse eliminates the linear momentum of fish P relative to fish Q.
The concept of the impulse as setting initial conditions is a powerful and commonly
used one. It is rare that the detailed description of an impulsive force is available;
normally we work only with its overall effect on the motion. The examples above are
vivid demonstrations of this principle. We return to linear impulses again in Chapter 6
when addressing collisions and impacts.
4.2 Angular Momentum and Angular Impulse
This section introduces angular momentum.Angular momentum is not a new physical
concept; it is a mathematical manipulation that provides both new insights and new
ways of solving problems. Although its value may not be obvious now, angular
momentum and its integrated form—the angular impulse—are enormously useful for
solving many classes of problems, particularly those involving extended rigid bodies.
In this chapter, we introduce the definition of the angular momentum of a particle and
use it to solve some important examples. We return to it repeatedly in later chapters,
as it is a key tool for understanding and modeling complex multiparticle systems.
The key thing to remember about angular momentum is that it depends on a reference
point. We often—but not always—select the reference point as the origin of an inertial
reference frame.
118
CHAPTER FOUR
rP/O
I
pP/O
I
I
I
pP/O
e2
e2
P
P
I
rP/O
O
hP/O
O
e1
e3
e1
e3
(a)
(b)
Figure 4.4 (a) Assume position rP/O and linear momentum IpP/O are confined to the plane
spanned by e1 and e2 . (b) Angular momentum IhP/O is a vector in the e3 direction formed by
applying the right-hand rule to the cross product rP/O × IpP/O .
4.2.1 Angular Momentum Relative to an Inertially Fixed Point
The angular momentum IhP/O represents the motion of particle P relative to reference
point O. We always explicitly include the reference point in our notation for angular
momentum.
Definition 4.2 The angular momentum IhP/O of a point P relative to point O
fixed in frame I is
rP/O × IpP/O = rP/O × mP IvP/O .
hP/O =
I
Definition 4.2 is illustrated in Figure 4.4b. In this section we assume that the reference
point O is the origin of an inertial frame.
What have we gained by defining angular momentum? Let’s look at its inertial rate
of change. Differentiating Definition 4.2 using the product rule results in
Id
I
hP/O
dt
Id I
vP/O
rP/O × mP IvP/O + rP/O × mP
dt
dt
I
= mP vP/O × IvP/O +rP/O × IaP/O .
=
Id
=0
Note that the first term in parentheses is zero, since the cross product of any vector
with itself is zero (see Appendix B). Using Newton’s second law to further simplify
the result, we obtain
Id dt
hP/O = mP rP/O × IaP/O = rP/O × mP IaP/O = rP/O × FP .
I
This expression yields a new definition.
LINEAR AND ANGULAR MOMENTUM OF A PARTICLE
119
Definition 4.3 The moment acting on point P relative to point O due to the net
force FP is
MP/O =
rP/O × FP .
The final result is the angular momentum form of Newton’s second law:
Id
dt
hP/O = MP/O .
I
(4.7)
We have chosen our words carefully here. There is no new physics in Eq. (4.7). The
result was obtained simply by using vector algebra and Newton’s second law. We are
still studying the motion of particles and how they behave according to Newton’s
laws. Using some new definitions, we derived a new formula that is enormously
useful—especially when considering multiple particles and rigid bodies. In fact, this
new statement of the second law can already make solving some problems easier, as
shown in the following examples.
Example 4.3 Calculating the Angular Momentum of a
Moving Point Mass
In this example we compute the angular momentum of a particle P with mass mP =
1 kg relative to the origin O, as shown in Figure 4.5. Using Cartesian coordinates, the
instantaneous position of the particle as components in frame I is
rP/O = 4e1 + 3e2 m.
The instantaneous velocity relative to I is also given as components in I :
I
vP/O = −6e1 + 8e2 m/s.
(4.8)
Using Definition 4.2, the instantaneous angular momentum of P relative to O is
I
hP/O = (1)(4e1 + 3e2) × (−6e1 + 8e2)
= 32(e1 × e2) − 18(e2 × e1)
= 32e3 + 18e3
= 50e3 kg-m2/s2.
It is also interesting to compute the angular momentum using the polar frame, also
shown in Figure 4.5. In terms of components in B the instantaneous position of P is
rP/O = 5er .
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CHAPTER FOUR
I
B
I
e2
vP/O
P
er
eθ
rP/O
3
e1
O
4
Figure 4.5 Point P located at Cartesian coordinates (4, 3)I in reference frame I traveling at
velocity IvP/O .
Referring to Section 3.4.1, we can write the instantaneous transformation array
between the two frames as
e1
e2
er
4
5
eθ
− 53
3
5
4
5
.
The instantaneous inertial velocity is then found from Eq. (4.8) and the transformation
array,
4
3
3
4
er − eθ + 8
e r + eθ
5
5
5
5
I
vP/O = −6
= 10eθ .
The instantaneous angular momentum is then again found from the definition:
I
hP/O = (1)(5er ) × (10eθ )
= 50(er × eθ )
= 50e3.
Example 4.4 Solving the Pendulum Using Angular Momentum
This example shows how using angular momentum makes the simple pendulum even
easier to solve. We again refer to Figure 3.18b, which illustrates the pendulum and
the polar frame B = (O, er , eθ , e3). As in Eq. (3.38), the position of the pendulum
bob is
rP/O = ler
and the velocity of the pendulum bob given in Eq. (3.39) is
I
vP/O =
Id
dt
(ler ) = l IωB × er = l θ̇eθ .
Using Definition 4.2, the angular momentum of the pendulum bob about the origin
O is
hP/O = rP/O × mP IvP/O = ler × mP l θ̇eθ = mP l 2θ̇e3.
I
LINEAR AND ANGULAR MOMENTUM OF A PARTICLE
121
Therefore, since e3 is fixed in I , we have
Id
dt
hP/O = mP l 2θ̈e3.
I
(4.9)
As before, the total force on the pendulum bob is FP = −F er + mP ge1. Using
er × er = 0, the moment of the pendulum bob about O is
MP/O = rP/O × FP = ler × mp g(cos θer − sin θeθ )
= −lmP g sin θe3.
(4.10)
Note that the tension force F does not contribute to the moment MP/O . Substituting
Eqs. (4.9) and (4.10) into Eq. (4.7) produces
mP l 2θ̈e3 = −lmP g sin θe3,
which can be solved to find the equation of motion
θ̈ +
g
sin θ = 0.
l
This approach is even more compact than the approach taken in Example 3.11. By
using angular momentum, we eliminated the force of constraint from the problem and
worked immediately with only a single equation for the single degree of freedom.
We cannot emphasize enough that the angular momentum of a particle depends on
a reference point. The following example reinforces this fact.
Example 4.5 Angular Momentum of a Particle Relative to
Two Different Points
This example compares the angular momentum IhP/O of particle P with respect to
point O , the origin of a polar frame B = (O , er , eθ , ez ), to the angular momentum
Ih
P/O of P with respect to O, the origin of an inertial frame I = (O, ex , ey , ez ).
Particle P moves along a circular trajectory centered at O with radius l, as shown in
Figure 4.6a. The vector rO /O is oriented along the ex axis and has (fixed) magnitude
d. The reference frames I and B, illustrated in Figure 4.6b, are related by the
transformation table
ex
er
eθ
cos θ
− sin θ
ey
sin θ
cos θ .
(4.11)
The kinematics of P with respect to O are
rP/O = ler
I
vP/O = l θ̇eθ .
Using Definition 4.2, the angular momentum of P with respect to O is
I
hP/O = ler × mP l θ̇eθ = mP l 2θ̇ez .
(4.12)
122
CHAPTER FOUR
P
l
d
O
θ
I
ey
O
O′
(a) Coordinates
B
eθ
er
ex
O′
(b) Reference frames
Figure 4.6 Angular momentum of a particle relative to two different points.
The kinematics of P with respect to O are
rP/O = dex + ler
I
vP/O = l θ̇eθ ,
where we used the fact that ḋ = 0. Using Definition 4.2 and the transformation table
Eq. (4.11), the angular momentum of P with respect to O is
I
hP/O = (dex + ler ) × mP l θ̇eθ = mP (dl θ̇ex × (− sin θex + cos θey ) + l 2θ̇ez )
= mP l θ̇ (d cos θ + l)ez .
(4.13)
Comparing Eqs. (4.12) and (4.13), we observe that IhP/O = IhP/O , as IhP/O
depends on the polar angle θ, whereas IhP/O does not.
We conclude this section with a discussion of angular momentum conservation.
Recall from Section 2.2.5 that Newton’s first law was equivalent to the law of conservation of linear momentum. In the absence of forces on a particle, the linear
momentum is a constant of the motion. We can state a similar law for angular momentum by noting from Eq. (4.7) that, in the absence of moments acting on a particle,
its angular momentum is a constant of the motion:
I
hP/O = constant.
This observation gives us a new law.
Law 4.1 The law of conservation of angular momentum of a particle states that,
when the total moment acting on a particle relative to point O is zero, the inertial
angular momentum of the particle relative to O is a constant of the motion.
It is important to note that angular momentum can be conserved even when there is
a nonzero force acting on a particle, as long as the total moment about O is zero. This is
in contrast to linear momentum, where the total force must be zero for the momentum
to be a constant. We discuss an important example of this in Section 4.2.4.
4.2.2 Angular Momentum Relative to an Arbitrary Point
Our discussion in the previous section is a special case of the more general scenario
in which the angular momentum is defined relative to an arbitrary point that may be
accelerating in absolute space.
LINEAR AND ANGULAR MOMENTUM OF A PARTICLE
123
I
vQ/O
B
B
Q
I
e2
rQ/O
vP/O
P
Q
I
I
e2
rP/Q
P
rP/O
O
I
I
vP/Q
hP/Q
O
e1
e3
e1
e3
(a)
(b)
Figure 4.7 Angular momentum relative to an arbitrary point. (a) Points P and Q move in
the plane spanned by e1 and e2 . There is a nonrotating frame B attached to Q. (b) Angular
momentum IhP/Q is the vector in the e3 direction formed by applying the right-hand rule to
the cross product rP/Q × mP IvP/Q, where IvP/Q is equal to IvP/O − IvQ/O .
Figure 4.7 illustrates the angular momentum of point P relative to an arbitrary
point Q moving in the inertial reference frame I . We also introduce a nonrotating but
translating frame B fixed to Q. Using Definition 4.2, the angular momentum relative
to Q is
hP/Q = rP/Q × mP IvP/Q,
I
(4.14)
I where IvP/Q = dtd rP/Q .1 We substitute in Eq. (4.14) for both the position and velocity
of P relative to Q using the appropriate vector triads from Figure 4.7 to obtain
I
hP/Q = (rP/O − rQ/O ) × mP (IvP/O − IvQ/O ).
The inertial rate of change of the angular momentum is
Id
dt
I
hP/Q
=
Id
dt
rP/O − rQ/O × mP IvP/O − IvQ/O
=0
Id
+ (rP/O − rQ/O ) × mP
= (rP/O − rQ/O ) ×mP
dt
I
vP/O − IvQ/O
aP/O − IaQ/O ,
I
=rP/Q
1 As in the discussion of relative motion in Chapter 3, we could replace the velocity Iv
P/Q in Eq. (4.14)
with BvP/Q , since the reference frame B is only translating and not rotating. However, subsequent chapters
consider both rotating and translating frames, and this notation would introduce much confusion. Therefore
we continue to use IvP/Q .
124
CHAPTER FOUR
where the first term is zero because a vector crossed into itself is zero. Finally, from
Newton’s second law, we substitute mP IaP/O = FP to find
Id
hP/Q = rP/Q × FP −rP/Q × mP IaQ/O .
I
dt
= MP/Q
The term MP/Q is the moment acting on P relative to Q. This calculation establishes the final expression for the inertial rate of change of the angular momentum of
P relative to Q:
Id
dt
hP/Q = MP/Q − rP/Q × mP IaQ/O ,
I
(4.15)
where
rP/Q × FP .
MP/Q =
Eq. (4.15) is the most general equation for the inertial rate of change of angular
momentum relative to an arbitrary point Q. By comparison to Eq. (4.7), observe
that writing the angular momentum form of Newton’s second law relative to an
accelerating point Q adds a correction term −rP/Q × mP IaQ/O , just as writing the
linear form in an accelerating frame added a similar correction term in Chapter 3,
Eq. (3.57). The correction term is sometimes called the inertial moment. Note also
that, if point Q is not accelerating, that is, if IaQ/O = 0, then frame B is an inertial
frame and Eq. (4.15) equals Eq. (4.7).
Example 4.6 Solving the Pendulum Using Angular Momentum in an
Accelerating Frame
Here we consider again the pendulum in an accelerating box of Example 3.14. This
time, however, we use angular momentum and Eq. (4.15) to solve for the equation of
motion. Figure 4.8 is the same as Figure 3.27 except that we have added an additional
polar frame C. The origin of C is O , the base of the pendulum.
Following Example 4.4, we write the position of the pendulum bob relative to O using polar coordinates as a vector component in the polar frame C,
rP/O = ler .
The corresponding velocity is
I
vP/O = l θ̇eθ .
The angular momentum of P relative to the (accelerating) attachment point O is
hP/O = rP/O × mP IvP/O = mP l 2θ̇e3.
I
Using Eq. (4.10), the moment on the pendulum bob about O is
MP/O = −lmP g sin θe3.
LINEAR AND ANGULAR MOMENTUM OF A PARTICLE
125
eθ
x
O′
O
O
e2
O′
C
b2
g
θ
l
e1
mP
a
er
b1
I
–Fer
B
mPge1
(a) Coordinates
(b) Reference frames
(c) Free-body diagram
Figure 4.8 Solving the pendulum problem using angular momentum in an accelerating frame.
Because the point O is accelerating at IaO /O = ae2, we use Eq. (4.15) for the
equation of motion. The inertial moment term in Eq. (4.15) is
rP/O × mP IaO /O = mP ler × ae2 = mP la cos θe3,
where we used rP/O − rO /O = rP/O and er × e2 = cos θe3 (see the transformation
table in Example 3.11).
Using the fact that e3 is fixed in I , Eq. (4.15) becomes
Id dt
hP/O = mP l 2θ̈e3 = −lmP g sin θe3 − lmP a cos θe3.
I
This vector equation is equivalent to the single scalar equation of motion
θ̈ +
a
g
sin θ = − cos θ.
l
l
(4.16)
Eq. (4.16) is the same equation of motion as in Eq. (3.58).
4.2.3 Angular Impulse
Just as we defined a linear impulse for short impulsive forces, we can define a
corresponding angular impulse when a moment acts only over a short interval.
Definition 4.4 The angular impulse MP/O (t1, t2) acting on particle P is the inte-
grated moment on P relative to point O over a (short) time interval t =
t2 − t1:
MP/O (t1, t2) =
t2
MP/O dt.
t1
As for the linear form of Newton’s second law, we compute the integral with respect
to time of the angular momentum form of Newton’s second law in Eq. (4.7),
I
hP/O (t2) = IhP/O (t1) + MP/O (t1, t2).
(4.17)
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CHAPTER FOUR
I
ey
P
P
B
l
eθ
θ
O
R
O
O′
O′
(a) Coordinates (top view)
–Ter
er
ex
(b) Reference frames
(c) Free-body diagram
Figure 4.9 Tetherball.
Our discussion in the previous section about linear impulses directly applies to angular
impulses and impulsive moments.
Example 4.7 Tetherball
Consider the (planar) tetherball apparatus shown in Figure 4.9. In this game, two
players hit the ball P in opposite directions and seek to be the first to wrap the tether
around the column centered at O. In this example we find the angular impulse that
reverses the velocity of the tetherball using a polar frame B = (O , er , eθ , ez ). Assume
the polar angle θ satisfies θ = 0 when the tether is completely unwrapped, which
implies that l = L − Rθ , where L is the total length of the tether and R is the radius
of the column. We ignore the thickness of the tether. The kinematics of the ball are
rP/O = −Reθ + ler
= −Reθ + (L − Rθ)er
I
vP/O = R θ̇er − R θ̇er + (L − Rθ)θ̇eθ
= (L − Rθ)θ̇eθ .
(4.18)
Using Definition 4.2, the angular momentum of P about O is
hP/O = −Reθ + (L − Rθ)er × mP (L − Rθ )θ̇eθ
I
= mP (L − Rθ)2θ̇ez .
To reverse the ball’s velocity by hitting it at time t1, we seek
I
hP/O (t2) = rP/O (t1) × mP (−IvP/O (t1)) = −IhP/O (t1),
where t2 is the time immediately after the impact. Using Eq. (4.17), we obtain
MP/O (t1, t2) = IhP/O (t2) − IhP/O (t1)
= −2IhP/O (t1) = −2mP (L − Rθ)2θ̇ez .
(4.19)
LINEAR AND ANGULAR MOMENTUM OF A PARTICLE
127
Next we find the equation of motion for θ after the impact. Using Definition 4.3
and Figure 4.9c, the moment on P about O after the impact is
MP/O = (−Reθ + ler ) × (−T er ) = −RT ez ,
(4.20)
where T is the tension in the tether. Eq. (4.20) implies that the angular momentum of
P about O is not conserved. Using Eq. (4.7), Eq. (4.20), and the time derivative of
Eq. (4.19), the equation of motion is
θ̈ =
RT
2R θ̇ 2
−
.
L − Rθ
mP (L − Rθ )2
(4.21)
But what is the tension T ? To find T , we use Newton’s second law again. Using
Figure 4.9c and the inertial derivative of Eq. (4.18), we have
−T er = −mP (L − Rθ)θ̇ 2er + mP (−R θ̇ 2 + (L − Rθ )θ̈)eθ .
(4.22)
Equating the er terms yields T = mP (L − Rθ)θ̇ 2. Thus the equation of motion in
Eq. (4.21) becomes
θ̈ =
m R(L − Rθ)θ̇ 2
R θ̇ 2
2R θ̇ 2
− P
=
.
L − Rθ
mP (L − Rθ )2
L − Rθ
(4.23)
(Note that the same equation of motion follows from the eθ term in Eq. (4.22).) We
illustrate the tetherball trajectory in Figure 4.10 by integrating Eq. (4.23) in matlab
using ODE45, with L = 10 m, R = 1 m, θ(0) = 0 rad, and θ̇(0) = 0.1 rad/s.
10
y (m)
5
0
–5
–10
–10
–5
0
5
10
x (m)
Figure 4.10 Trajectory of the tetherball as it winds around the post after an impact. (The rope
is not shown.)
128
CHAPTER FOUR
4.2.4 Central-Force Problems
The angular momentum form of Newton’s second law in Eq. (4.7) and the law of
conservation of angular momentum reveal an important insight into a broad class
of problems with so-called central forces. The concept of a central force, and the use
of angular momentum in solving a central-force problem, is so important that we
devote a section here to introducing it.
Definition 4.5 A central force is a force FP acting on particle P that satisfies the
relationship
MP/O = rP/O × FP = 0,
where rP/O = 0.
Definition 4.5 implies that the force FP acts parallel to the line between the origin
O and the particle P . An important consequence of Definition 4.5 is that the angular
momentum of P about O is conserved since it satisfies the zero-moment condition
in the law of conservation of angular momentum of a particle (as long as O is not
accelerating in absolute space).
The time invariance of the angular momentum implies that the motion of point
P is confined to a plane perpendicular to the angular momentum—an implication of
Definition 4.2 of angular momentum. Since IhP/O = rP/O × mP IvP/O , and IhP/O is
fixed in the inertial frame, then rP/O and IvP/O must be confined to a plane so that they
complete a right-hand set with the fixed IhP/O . We call the motion of P in the plane
perpendicular to IhP/O an orbit. We can thus reduce any central-force problem to a
two-dimensional particle dynamics problem similar to the ones solved in Chapter 3.
This reduction is a profound result. Also, because of the radial form of the central
force, we often solve such problems using polar coordinates in a polar frame.
Example 4.8 A Particle on a Planar Spring, Revisited
Example 3.12 considered the motion of particle P of mass mP connected to a linear
spring and free to move anywhere in the plane. We described this two-degree-offreedom problem with polar coordinates and found the two scalar equations of motion
to be
r̈ +
k
(r − r0) − r θ̇ 2 = 0
mP
ṙ
θ̈ + 2 θ̇ = 0.
r
This is an example of a central-force problem, as the only force acting on P is due
to the spring and always acts in the er direction. Thus the trajectory of P is an orbit in
the plane determined by these equations of motion. We also know that the angular momentum of the particle relative to the origin O must be conserved. This angular
momentum is
LINEAR AND ANGULAR MOMENTUM OF A PARTICLE
129
hP/O = rP/O × mP IvP/O
I
= rer × mP (ṙer + r θ̇eθ )
= mP r 2θ̇e3.
Since the magnitude and direction of the angular momentum are fixed, we use the
constant hO to represent the specific magnitude of the angular momentum:
hO =
IhP/O mP
= r 2θ̇.
This constant is the same found in Example 3.12 (Eq. (3.48)) simply by reducing the
equations of motion! Thus we can make a substitution in the equations of motion,
leaving us with what we found in Example 3.12,
r̈ +
h2
k
(r − r0) − O3 = 0.
mP
r
(4.24)
This result is remarkable and important. Because of the conservation of angular
momentum, we were able to decouple the radial and angular equations of motion.
Given a set of initial conditions that determine hO , we need only integrate the single
second-order differential equation for r in Eq. (4.24). The angular position θ can then
be found simply from the constancy of angular momentum,
t
hO
θ(t) =
dt + θ(0).
2
0 r(t)
This is true for all central force problems.
Example 4.9 Simple Satellite Equations of Motion
The most important central force we will study is gravity—described by Newton’s
universal law of gravitation (see Tutorial 2.5), which in vector form is
FP = −
GmO mP
rP/O .
rP/O 3
(4.25)
Here we have assumed a gravitating body of mass mO located at the fixed origin
O and a corresponding satellite P of mass mP at rP/O , as shown in Figure 4.11.2
Eq. (4.25) allows us to write the vector form of the satellite equation of motion, using
the free-body diagram in Figure 4.11c and Newton’s second law:
Id 2
dt 2
rP/O +
GmO
rP/O = 0.
rP/O 3
(4.26)
2 A model with a fixed origin does not exactly represent two bodies in space, since the sun—or the earth—
also moves. It is, however, an exact description of a central-force problem with a fixed attracting origin and
is a remarkably good approximation of a real satellite orbit problem. Chapter 7 treats the exact two-body
problem more carefully.
130
CHAPTER FOUR
I
I
e2
P
P
B
r
er
eθ
θ
O
O
(a) Coordinates
GmOmP
–—
er
r2
e1
(b) Reference frames
(c) Free-body diagram
Figure 4.11 A simple satellite.
Although Eq. (4.26) looks like a formidable equation of motion, it turns out
to be solvable. One nice way to approach it is by using polar coordinates. Let
B = (O, er , eθ , ez ) be a polar frame, where er points from O to P , and er and eθ
span the orbit plane. (Because of the constancy of angular momentum, we know that
the orbit is confined to a plane and that we can treat the motion in two dimensions.)
The kinematics of P expressed as vector components in frame B are
rP/O = rer
I
vP/O = ṙer + r θ̇ eθ
I
aP/O = (r̈ − r θ̇ 2)er + (2ṙ θ̇ + r θ̈)eθ .
Again using the free-body diagram in Figure 4.11c and the gravitational force in
Eq. (4.25), we write Newton’s second law in Eq. (4.26) in terms of components in B,
−
GmO mP
er = mP (r̈ − r θ̇ 2)er + mP (2ṙ θ̇ + r θ̈)eθ .
r2
This results in two scalar equations of motion
2ṙ θ̇
r
GmO
r̈ = − 2 + r θ̇ 2.
r
θ̈ = −
(4.27)
(4.28)
We can now use the conservation of angular momentum to reduce the two equations
of motion for the polar coordinates (r, θ)I to a single differential equation for r. Since
the angular momentum is constant, it is determined by the initial radial position and
initial velocity—these initial conditions thus determine the characteristics of the orbit.
Using Definition 4.2, the angular momentum of satellite P about origin O is
I
hP/O = rer × mP (ṙer + r θ̇ eθ ) = mP r 2θ̇ez .
Again, since the magnitude and direction of the angular momentum are fixed, we use
the constant hO to represent the specific magnitude of the angular momentum:
hO =
IhP/O mP
= r 2θ̇.
LINEAR AND ANGULAR MOMENTUM OF A PARTICLE
131
Solving for θ̇ = hO /r 2 and substituting this expression into the satellite equations of
motion in Eqs. (4.27) and (4.28) yields
θ̈ = −
2ṙhO
r3
(4.29)
r̈ = −
h2O
GmO
+
.
r2
r3
(4.30)
Eqs. (4.29) and (4.30) are significant because the equation of motion in r is
uncoupled from the equation of motion in θ. In fact, Eq. (4.30) can be analytically
integrated (using a surprisingly simple change of coordinates) to find the satellite
trajectory r(θ ) (see Tutorial 4.2). Even without integrating, Eqs. (4.29) and (4.30)
demonstrate the existence of closed, circular orbits about O. If we consider the steadystate condition r̈ = θ̈ = 0 for circular orbit, the equations of motion reduce to
GmO
θ̇ =
r03
ṙ = 0,
which is a circular orbit with radius r0. Using the kinematics of P , the satellite velocity
in a circular orbit of radius r0 is
GmO
I
vP/O =
eθ .
r0
It is also interesting to point out another special trajectory. If the initial velocity
is zero, then hO = 0 and there is no angular motion. In this case, the particle simply
accelerates on a linear trajectory toward the origin, as in Tutorial 2.5.
4.3 Tutorials
Tutorial 4.1 A Passive Walker
A recent discovery in biomechanics is that robotic bipeds can stably walk down an
incline—that is, walk without falling—without any control of the hips or legs from
the brain. Walking without active control of the hips or legs is generally referred to
as passive dynamic walking and has important implications in the study of neuromechanics. With only a few simplifying assumptions, the tools of this chapter can be
used to model and study this remarkable result.
Figure 4.12 shows a simple model of a passive walking robot. We model the robot
using a point mass P to represent the body and two massless rods to represent the
legs. The robot “walks” down a ramp by tipping forward, like an inverted pendulum,
and transferring its weight from one leg to the other. The point of contact of each leg
with the ramp is called a foot. Each stride starts and ends with both feet on the ramp.
We assume that the dynamics essential to walking are confined to a plane.
Let l denote the length of the legs and assume that every stride has a fixed length.
The angle 0 < α < π denotes the angular separation of the legs at the instant when both
132
CHAPTER FOUR
P
g
l
β
α
I
θ′
O′
B′
θ′
O′
O″
B″
O″
N′er ′
er ′
ey
–mPgey
eθ ′
O
(a) Coordinates
ex
(b) Reference frames
(c) Free-body diagram
Figure 4.12 A passive walker. The coordinates, reference frames, and free-body diagram are
shown for a single stride. Stable walking consists of an infinite series of strides.
feet are on the ramp.3 Let 0 < β < π2 denote the ramp inclination angle. Intuition tells
us that, if the ramp is not steep enough, then the robot’s momentum may be insufficient
to sustain continuous motion down the ramp, in which case the robot falls backward.
Alternatively, if the ramp is too steep, then the robot may gain too much momentum
and lose contact with (or topple off ) the ramp. In this tutorial we model the walking
dynamics using linear impulse and, for a given α, find the sufficient conditions on β
that support walking.
Equations of Motion
Let I = (O, ex , ey , ez ) denote an inertial frame, as shown in Figure 4.12b. Let
B = (O , er , eθ , ez ) denote the polar frame fixed to one leg with origin O located at
the foot of the leg. The corresponding angle θ is the angle of the leg relative to vertical.
Let B = (O , er , eθ , ez ) and θ denote the polar frame and angle associated with
the other leg. The three reference frames are related by
ey
ex
er eθ θ
− sin
− cos θ er cos θ − sin
θ
cos α
− sin α
eθ sin α
cos α ,
where the transformation between B and B is valid only when both feet are touching
the ramp (when the angle between the legs is α).
Consider, for the moment, the geometry of the robot with both feet on flat ground,
that is, with β = 0. In this configuration, the angle between the “back” leg and ey is
α/2, whereas the angle between the “front” leg and ey is −α/2. If the robot is walking
3 We do not model the mechanism at the hip joint that controls the leg placement. Our model is equivalent
to assuming the robot has 2π/α massless legs uniformly splayed in all directions, so that the robot “rolls”
bumpily down the ramp or, equivalently, that the rear leg instantaneously rotates to the front after leaving
the ramp. In a more detailed model we might include the mass of the feet and treat the legs as pendulums.
LINEAR AND ANGULAR MOMENTUM OF A PARTICLE
133
1 –
— FP(t1,t2)
mP
t = t0
t = t1
α
β+–
2
I
vP/O″ (t2)
α
β––
2
α
I
vP/O′ (t1)
O′
O′
O″
β
(a)
(b)
Figure 4.13 (a) Front leg during the duration of a single stride. The stride starts at t = t0 when
θ = β − α/2 and the rear leg has just lost contact with the ramp. The stride ends at t = t1
when θ = β + α/2 and the new front leg has just made contact with the ramp. (b) As the new
front foot is planted, a linear impulse redirects the velocity to be tangent to a circular trajectory
centered at O .
stably down an inclined ramp, then a stride supported by the first leg starts at time
t = t0 at an angle of
θ (t0) = β −
α
2
(4.31)
θ (t1) = β +
α
.
2
(4.32)
and ends at t = t1 at the angle
Such a stride is shown in Figure 4.13a.
During a stride, the kinematics of P are
rP/O = ler I
vP/O = l θ̇ eθ aP/O = l θ̈ eθ − l(θ̇ )2er .
I
Let N denote the magnitude of the normal force from the ramp during the stride.
Using the free-body diagram in Figure 4.12c, the dynamics of P are
N er − mP gey = mP l θ̈ eθ − mP l(θ̇ )2er ,
which yields the force equation
N = mP g cos θ − mP l(θ̇ )2
(4.33)
and the equation of motion
θ̈ =
g
sin θ .
l
(4.34)
134
CHAPTER FOUR
Eq. (4.34) is the equation of motion for an inverted pendulum (see Problem 4.3) and
is the same equation of motion as in Tutorial 3.3. (Note the sign change from the
equation of motion for the simple pendulum.)
For the robot to remain in contact with the ramp throughout the stride and to
not topple off, the normal force N must satisfy N > 0 or, equivalently, the specific
normal force N /mP must satisfy (N /mP ) > 0. We call this inequality the toppling
condition.
Impulsive Stride Transfer
During a stride, the velocity IvP/O is in the eθ direction, which implies that the
trajectory of P is tangent to a circle with radius l centered at O . When the stride ends
and the next stride begins, the new velocity IvP/O is in the eθ direction. Therefore
the velocity of P changes direction from eθ to eθ during the short time interval
t = t2 − t1 when both feet are in contact with the ramp. This situation is illustrated
in Figure 4.13b.
We study the change in the velocity using the impulse equation
mP IvP/O (t2) = mP IvP/O (t1) + mP IvO /O (t1) + FP (t1, t2),
=0
which, after substituting the kinematics, becomes
mP l θ̇ (t2)eθ = mP l θ̇ (t1)eθ + FP (t1, t2).
(4.35)
eθ = − sin αer + cos αeθ (4.36)
The relationship
is satisfied during the time interval t1 to t2. Using Eq. (4.36), we compute the linear
impulse that eliminates the er component of the velocity:
FP (t1, t2) = mP l θ̇ (t1) sin αer .
(4.37)
We find the impulse magnitude FP (t1, t2) by solving the equation of motion during
the stride for the final angular velocity. Substituting Eq. (4.37) into Eq. (4.35) and
using Eq. (4.36) yields
θ̇ (t2) = θ̇ (t1) cos α.
(4.38)
Eq. (4.38) provides the initial condition for the second (nth) stride in terms of the final
condition of the first ((n − 1)th) stride.
Sufficient Conditions for Walking
A sufficient condition for walking expressed in terms of the parameters α, β, and l
can be derived from
θ̇ (t2) = θ̇ (t0).
(4.39)
LINEAR AND ANGULAR MOMENTUM OF A PARTICLE
135
This condition states that the angular momentum of P about the contact point at the
start of each stride is the same.4 Let h0 denote the magnitude of the specific angular
momentum at the start of each stride:
h0 =
1 I
hP/O (t0) = ler × l θ̇ (t0)eθ = l 2θ̇ (t0).
mP
(4.40)
Multiplying Eq. (4.34) by θ̇ dt and integrating from t0 to t0 < t ≤ t1 yields
θ̇ (t)
θ̇ (t0 )
g
θ̇d θ̇ =
l
θ (t)
θ (t0 )
sin θdθ,
which, using Eqs. (4.31) and (4.40), implies
2g
(cos(θ (t0)) − cos(θ (t)))
θ̇ (t) = (θ̇ (t0))2 +
l
h20 2g
α
+
cos β −
− cos(θ (t)) .
=
4
l
l
2
(4.41)
When t = t1 (the end time of the stride), we find, using Eqs. (4.38) and (4.41), that
h2 2g
α
α
θ̇ (t2) = cos α 40 +
cos β −
− cos β +
l
l
2
2
h2 4g
α
= cos α 40 +
(4.42)
sin β sin ,
2
l
l
where we also used Eqs. (4.31) and (4.32) and the trigonometric identity cos(β ± α2 ) =
cos β cos α2 ∓ sin β sin α2 . We now substitute for θ̇ (t2) from the walking condition
in Eq. (4.39) and replace θ̇ (t0) by the angular momentum in Eq. (4.40) to obtain
h20 4g
h0
α
+
=
cos
α
sin β sin .
2
l4
l
l2
After some algebra, this expression simplifies to
α
h0 = 2l cot α gl sin β sin .
2
(4.43)
Suppose t ∗ satisfies θ (t ∗) = 0, which is the time that P is vertical. For the robot to
tip forward and not backward, we require that θ̇ (t ∗) > 0. Using Eq. (4.41), a falling
condition is then
h20
l4
+
2g
l
cos β −
α
2
− 1 > 0,
4 Another mode of walking is, for example, when the angular momentum of P about the contact point at
the start of alternate strides is the same.
136
CHAPTER FOUR
90
80
Leg angle α (°)
70
60
β = 10.2°
Fall
β = 15.3°
50
40
30
20
Topple
10
0
0
5
10
15
20
Ramp angle β (°)
Figure 4.14 A passive walker. Values of α and β shown in the shaded region are sufficient but
not necessary for stable walking with l = 1 m. For example, for α = 60◦, the robot falls when
β < 10.2◦ and topples when β > 15.3◦.
where we have used cos(θ (t ∗)) = 1. Using Eq. (4.43), this inequality is equivalent to
2 cot 2 α sin β sin
α
α
+ cos(β − ) − 1 > 0.
2
2
(4.44)
If the robot doesn’t fall, then the angular momentum of P about the contact point
is maximum just before the footfall at time t1. Thus from Eq. (4.33) the normal force
N is a minimum at t1. Evaluating Eq. (4.33) at t1 using Eqs. (4.32) and (4.38), the
toppling condition becomes
g cos θ (t1) − l(θ̇ (t1))2 = g cos(β +
α
(θ̇ (t2))2
> 0,
)−l
2
cos2 α
which, using Eqs. (4.39) and (4.43), is equivalent to
cos(β +
h2
4 sin β sin
α
α
) − 3 0 2 = cos(β + ) −
2
gl cos α
2
sin2 α
α
2
> 0.
(4.45)
Using Eqs. (4.45) and (4.44), we plot in Figure 4.14 the values of α and β that, for
l = 1 m, are sufficient for walking. Interestingly, these values are independent of g,
which means that these results are independent of the acceleration of gravity.
Tutorial 4.2 The Orbit Equation
This tutorial picks up where we left off in Example 4.9 and derives the general
equation for a satellite orbit. Recall the observation made in Example 4.9. The angular
momentum of satellite P about origin O is constant, since the only force acting on
P is gravity—which is a central force. The magnitude and direction of the angular
LINEAR AND ANGULAR MOMENTUM OF A PARTICLE
137
momentum are constants determined by the initial position and initial velocity of P .
We denoted by hO the specific magnitude of the angular momentum:
hO =
IhP/O mP
= r 2θ̇.
The satellite equations of motion (Eqs. (4.29) and (4.30)) are
θ̈ = −
2ṙhO
r3
(4.46)
r̈ = −
h2O
μ
+
,
r2
r3
(4.47)
GmO .
where μ =
Unfortunately, there is no known solution of Eqs. (4.46) and (4.47) for the two
remaining degrees of freedom, r(t) and θ(t), in terms of elementary functions. This
is a well-known problem in celestial mechanics referred to as Kepler’s problem. It
is beyond our scope to discuss Kepler’s problem. However, with a simple change of
coordinates, we can integrate Eq. (4.47) to find an implicit expression for the satellite
trajectory r(θ ).
Let y = 1/r. Then we have hO = θ̇/y 2 and
ṙ =
d
dt
1
y
=−
1
1 dy
dy
ẏ = − 2
θ̇ = −hO ,
2
y
y dθ
dθ
which implies
r̈ = −hO
2
d 2y
2 2d y
.
θ̇
=
−h
y
O
dθ 2
dθ 2
(4.48)
Substituting Eq. (4.48) in Eq. (4.47) and switching to the y coordinate yields
d 2y
μ
+y= 2 .
2
dθ
hO
(4.49)
Eq. (4.49) is the equation for a simple harmonic oscillator with unit oscillation
frequency and constant driving force! Tutorial 2.3 showed that the solution to this
equation is sinusoidal about a constant offset, that is,
μ
+ A cos θ + B sin θ.
h2O
y(θ) =
For this problem it is more convenient to write the solution in terms of an amplitude
and phase rather than the integration constants A and B (in both cases the constants
are given by initial conditions):
y(θ) =
μ
μ
+ 2 e cos(θ − θ0),
h2O
hO
where
e=
h2O μ
A2 + B 2
138
CHAPTER FOUR
Table 4.1
The orbit shape corresponding to
eccentricity e
Eccentricity e
Orbit shape
e=0
circle
0<e<1
ellipse
e=1
parabola
e>1
hyperbola
and
θ0 = arctan
−B
A
.
Therefore, in terms of the r coordinate, we have
r(θ) =
h2O /μ
1 + e cos(θ − θ0)
.
(4.50)
Eq. (4.50) is the polar equation of a conic section5 with eccentricity e. It is not
difficult to show that the eccentricity is nonnegative. The dependence of orbit shape
on eccentricity is described in Table 4.1.
Many orbit problems can be solved using Eq. (4.50). The satellite’s radial position
r for any given angular position θ follows a conic section determined by e with the
gravitating body at one focus, as shown in Figure 4.15. The specific size and shape of
the orbit is given by the initial conditions. That is, if we consider the radial position
and angular velocity at the reference angle θ0, polar radius r(θ0), and polar angle rate
θ̇ (θ0), we have
hO = r 2(θ0)θ̇(θ0)
e=
r 3(θ
0
)θ̇ 2(θ
μ
0)
(4.51)
− 1.
(4.52)
Eq. (4.50) shows that the polar radius is smallest when θ = θ0. Thus the angle
reference θ0 corresponds to the point of closest approach—known as periapsis. The
radius at this point is
r(θ0) =
rp =
h2O
μ(1 + e)
.
(4.53)
5 A conic section is the intersection of a plane with a cone. Depending on the intersection angle, the section
is a circle, an ellipse, a parabola, or a hyperbola.
LINEAR AND ANGULAR MOMENTUM OF A PARTICLE
139
3
e=0
e = 0.6
e=1
e = 1.5
2
1
0
–1
–2
ra = 2.5
–3
–3
–2
rp = 0.625
–1
0
1
2
3
Figure 4.15 Eq. (4.50) with h2O = μ plotted for three values of eccentricity e. For e = 0.6,
the trajectory is an ellipse; the periapsis rp and apoapsis ra denote the closest and farthest,
respectively, points of approach to the origin at (0, 0).
The farthest point of approach—known as apoapsis—corresponds to θ = θ0 + π,
which implies
r(θ0 + π) =
ra =
h2O
μ(1 − e)
.
(4.54)
Finally, the semimajor axis a of an ellipse (or circle) is defined as half the length of
the long axis:
h2O
1
a = (ra + rp ) =
.
μ(1 − e2)
2
(4.55)
Tutorial 4.3 Impulsive Orbital Transfer
The concept of angular impulse is used regularly in satellite orbital transfer. Consider
the situation sketched in Figure 4.16 in which a satellite is in a circular orbit of radius
rA and needs to be transferred to a larger circular orbit of radius rB (e.g., the satellite
may have been left in a low-earth orbit by a launch vehicle). To do so, the satellite
changes the eccentricity of its orbit to achieve a transfer orbit—an elliptical orbit with
periapsis equal to rA and apoapsis equal to rB .
For the satellite to move from the initial circular orbit to the eccentric transfer
orbit, it must change its angular momentum and eccentricity, given by Eqs. (4.51)
and (4.52), respectively. Satellites are normally equipped with propulsion devices that
can produce large forces over short intervals. The firing time is so brief compared to
the orbit period that it is very accurately modeled as an impulsive force. (We call
140
CHAPTER FOUR
––
MB/O
B
rB
O
rA
––
MA/O
A
Figure 4.16 Sketch of a Hohmann transfer orbit between two circular orbits of radii rA and
rB . Two angular impulses are required, one of M A/O at periapsis of the elliptical transfer orbit
and one of M B/O at apoapsis.
the resulting change in velocity a v, pronounced “delta-vee.”) The impulsive force
produces an impulsive moment about O. According to Eq. (4.17), a thrust applied at
time t1 and ending at time t2 produces a change in angular momentum of
I
hP/O (t2) = IhP/O (t1) + MP/O (t1, t2).
In the initial circular orbit, the satellite’s angular velocity is
θ̇A =
μ
.
rA3
Thus the specific angular momentum before the impulse has magnitude hO (t1) =
√
μrA. We wish the satellite to enter into the elliptic transfer orbit with periapsis rA
and apoapsis rB . From Eqs. (4.53) and (4.54) we find that the magnitude of the specific
angular momentum of the transfer orbit (i.e., after the impulse) is
hO (t2) =
2μrArB
.
rA + r B
(4.56)
The eccentricity of the transfer orbit is also found from Eqs. (4.53) and (4.54),
e=
rB − rA
.
rB + r A
Because the transfer orbit is tangent to the circular orbit at periapsis, the velocity
after the impulse is in the same direction as the velocity before the impulse, implying
that the impulsive force is tangent to both orbits. The magnitude of the angular impulse
required at point P = A is
LINEAR AND ANGULAR MOMENTUM OF A PARTICLE
MA/O (t1, t2) = hO (t2) − hO (t1) =
√
141
μrA
2rB
−1 .
rA + r B
Once the satellite reaches the apoapsis of the transfer orbit at radius rB , another
impulsive firing is required to “circularize” the orbit—that is, to change the angular
momentum to that of the final circular orbit. Switching from an elliptical orbit to a
circular orbit requires the satellite to speed up. The angular momentum before this
second impulse has magnitude hO in Eq. (4.56). The angular momentum after the
second impulse is equal to the angular momentum of the final circular orbit, which has
√
magnitude hO (t2) = μrB . The required impulsive moment at point P = B (again
from a force tangent to both orbits) is the difference between these angular momenta,
√
2rA
MB/O (t1, t2) = μrB 1 −
.
rA + r B
An orbital transfer using two impulses is by far the most commonly used and is
called a Hohmann transfer.6 A Hohmann transfer requires the least fuel of all possible
two-impulse transfers.
4.4 Key Ideas
.
Integrating Newton’s second law with respect to time yields
mP IvP/O (t2) = mP IvP/O (t1) + FP (t1, t2),
where FP (t1, t2) is the linear impulse acting on the particle:
t2
FP dt.
FP (t1, t2) =
t1
t2 − t1, a linear impulse changes the linear momentum
For sufficiently small t =
of a particle without changing its position.
.
The angular momentum of a particle relative to point O is
rP/O × mP IvP/O .
hP/O =
I
Angular momentum always depends on a reference point. The rate of change of
angular momentum when the reference point is inertially fixed is
Id dt
hP/O = MP/O ,
I
where the moment acting on P due to the force FP is
MP/O =
rP/O × FP .
6 Named after Walter Hohmann (1880–1945), a German scientist, who first described this transfer in 1925.
142
CHAPTER FOUR
.
More generally, the rate of change of angular momentum when taken relative to
an arbitrary point Q is
Id dt
hP/Q = MP/Q − rP/Q × mP IaQ/O ,
I
where IaQ/O is the inertial acceleration of the reference point.
.
Integrating with respect to time the angular momentum form of Newton’s second
law (referenced to an inertially fixed point) yields
I
hP/O (t2) = IhP/O (t1) + MP/O (t1, t2),
where MP/O (t1, t2) is the angular impulse,
MP/O (t1, t2) =
t2
MP/O dt.
t1
.
.
When the total moment acting on P relative to point Q (or O) is zero, then the
angular momentum of P relative to that point is conserved, even if there are forces
acting on P . The angular momentum is constant in both magnitude and direction.
A central force on P results in zero moment about O:
MP/O = rP/O × FP = 0.
Under the action of a central force, the angular momentum of a particle is conserved
and its motion is described by a two-dimensional orbit in a plane perpendicular to
the angular momentum.
4.5 Notes and Further Reading
In this chapter we begin to see the extension of mechanics beyond the simple statements of Newton’s laws. In fact, it was d’Alembert in 1742 who introduced the
concept of the impulse and Euler who introduced angular momentum. The concepts of
inertial forces and inertial moments were also introduced by Euler in his laws of mechanics and used by d’Alembert in his treatise. Details can be found in Dugas (1988)
and Truesdell (1968).
Newton’s formulation of the law of gravity—and his combining of it with his laws
of motion to successfully solve for the motion of celestial bodies—was the greatest
triumph of his laws of mechanics. (In 1705 Edmund Halley used them to correctly
predict the return of the comet named after him, thus sealing the case for Newton’s
laws.) A rather nice, concise history is in Vallado (2001). A more thorough and popular
treatment is in Peterson (1993). Our solution to the motion of a body in orbit under
gravity is just a brief introduction. The subject of orbital motion is rich and complex,
with many excellent texts devoted solely to it. We like many of them, including
LINEAR AND ANGULAR MOMENTUM OF A PARTICLE
143
Bate et al. (1971), Chobotov (1991), Kaplan (1976), and Vallado (2001). If interested
in learning more about orbital maneuvers, such as the Hohmann transfer, you might
consult Kaplan (1976) or Vallado (2001).
4.6 Problems
4.1
Let P represent mass m shown in Figure 4.17, which is traveling at IvP/O =
ẋex + ẏey , where ẋ > ẏ > 0. Point Q is fixed with respect to O, and rQ/O =
aex + bey .
I
ey
Q
P
Iv
P/O
O
ex
Figure 4.17 Problem 4.1.
a. Using Cartesian coordinates (x, y, z)I , compute the following vector
quantities: IpP/O , IhP/O , IpP/Q, and IhP/Q.
b. True or false: At the instant shown in Figure 4.17, the vectors IhP/O
and IhP/Q point in the same direction.
4.2
Consider a 104 kg spaceship in deep space. Suppose the spaceship’s thruster
produces 100 N of constant thrust. If the spaceship is moving at 50 m/s and
the thruster is anti-aligned with the spaceship’s velocity, how long should the
thruster fire to bring the spaceship to rest with respect to absolute space?
4.3
Find the equation of motion for the inverted simple pendulum shown in Figure
4.18. How does it differ from the equation for the simple pendulum?
mP
θ
g
l
O
Figure 4.18 Problem 4.3.
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CHAPTER FOUR
4.4
Solve the equation of motion for the inverted simple pendulum from the
previous problem for small deviations from vertical. How does the solution
differ from that for the simple pendulum (i.e., simple harmonic motion)?
4.5
A hockey player hits a puck of mass mP into a semicircular groove in the
ice of radius R. The ice has coefficient of Coulomb friction μc . What is the
impulse that must be imparted to the puck at the entrance to the groove such
that it makes it around the semicircle exactly once before stopping?
4.6
Suppose frame B = (O , b1, b2, b3) is traveling in the e1 direction at a constant
speed of v0 with respect to stationary frame I = (O, e1, e2, e3), as shown in
Figure 4.19. Mass mP is connected to point O by a spring with spring constant
k and rest length r0. The spring can freely pivot about O . Assume that the
positions of O and O are the same at time t = 0.
B
I
e2
b2
mP
P
k
O
e1
O′
v0
b1
Figure 4.19 Problem 4.6.
a. Using the coordinates of your choice, find the position rP/O and
velocity IvP/O of the mass with respect to O in I . [HINT: Introduce
a polar frame at O .]
b. Find the position rP/O and velocity IvP/O of the mass with respect to
O in I .
c. Draw a free-body diagram for mass mP . (There is no gravity in this
problem.)
d. Find the angular momentum IhP/O of the mass with respect to O in
I.
e. Show that the angular momentum of the mass with respect to O in I
is conserved, but the angular momentum with respect to O in I is not.
4.7
Figure 4.20 shows a simple model for the gravitational attraction of particle
P to an object with nonuniform density. Here, rather than a single mass at the
origin O, we have a dipole close to the origin separated by 2d. Assuming that
the motion of P starts in the plane of the dipole, then to a good approximation
(for r
d) the force acting on the particle due to the dipole is
FP ≈ −
2mP μ 3mP μd 2
3mP μd 2 sin(2θ)
2
+
(3
cos
θ
−
1)
e
−
eθ ,
r
r2
r4
r4
where μ = Gm.
LINEAR AND ANGULAR MOMENTUM OF A PARTICLE
145
P
r
θ
m
O
m
d
d
Figure 4.20 Problem 4.7.
Determine whether the angular momentum of P relative to O is conserved
for this problem. Is the motion still a planar orbit, as it was for a single attracting particle? Write down the resulting equations of motion for P . Numerically
compute the trajectory of P for the following parameters: r0 = 1.5625 distance
units, ṙ0 = 0, θ0 = 0.6435 radians, θ̇0 = 0.6144 radians/time unit, μ = 0.5 distance units3/time units2, and d = 0.3125 distance units.7 You should integrate
for 100 time units. Also compute the trajectory for the central force due to a
mass of 2m located at the origin and compare. Use the same initial conditions
for both integrations. Note that distance is measured in earth radii and time is
measured in fractions of an orbital period. [HINT: Numerical integration errors may cause the central-force trajectory to appear like a spiral rather than a
closed orbit. Try reducing the error tolerances used by ODE45 using ODESET.]
4.8
Consider again Problem 3.20 from Chapter 3. What is the magnitude and
direction of the angular momentum of the collar relative to the point O? Is it
conserved? Explain.
4.9
Show that the eccentricity of an orbit in terms of the polar coordinates of the
initial conditions is given by Eq. (4.52).
4.10 The International Space Station is in a circular orbit with an altitude of 250 mi
(400 km). It has a mass of 1.8 × 105 kg. Here are some constants you may need
to know:
μearth = 3.986 × 105 km3/s2
Rearth = 6,378 km.
a. What is the velocity of the space station?
b. What is the orbital period of the space station?
c. If a crew-rescue vehicle leaves the station, how much must its velocity
be changed so that its new orbit has its closest approach to the earth at
Rearth ? Ignore aerodynamic drag.
7 The
reason that d is not much, much less than r in this problem is that if more realistic numbers were
used, it would take far more orbits than we could reasonably expect you to simulate to see any noticeable
difference.
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CHAPTER FOUR
mP
θ
g
l
O
u(t)
Figure 4.21 Problem 4.11
4.11 Consider again the inverted pendulum from Problem 4.3. This time the base
can move back and forth at a known displacement u(t), as shown in Figure 4.21.
a. Using angular momentum, find the equations of motion including the
displacement u(t) of the base.
b. Suppose you are holding the bottom of the stick and can move it back
and forth with a motion given by u(t) to try to keep it balanced. If the
pendulum starts at an offset angle θ0, show that it can be returned to
vertical if you accelerate your hand with the following function:
μ=−
g sin θ + kθ + bθ̇
cos θ
for some constants k > 0 and b > 0. This is called a feedback law, as it
depends on the instantaneous angle of the pendulum. How might you
choose k and b?
4.12 A particle slides freely on a flat surface, attached to a massless string that
passes through a hole in the surface. At a given moment, the particle is moving
in a circle of radius r0 with angular velocity ω0. Assume that the string is pulled
through the hole at a constant rate. Find an expression for the tension in the
string as a function of the length r of string left on the surface.
4.13 A particle of mass m is attached to a spring, as in Figure 4.22. Initially the
spring is unstretched with length l0. An impulse is applied to give the particle
θ
l0
O
v0
k
45°
m
Figure 4.22 Problem 4.13
LINEAR AND ANGULAR MOMENTUM OF A PARTICLE
147
an initial velocity of magnitude v0 in the direction shown in the figure. During
the particle’s trajectory, the spring stretches to a maximum length of 4l0/3.
Assume there is no gravity acting on the system.
a. What is the impulse about O applied to the mass to give it the initial
conditions specified?
b. Find the equations of motion of the mass as a function of l and θ.
c. Solve for the spring stiffness k as a function of m, l0, and v0.
4.14 Reproduce Figure 4.14 in Tutorial 4.1 for five different values of the leg length
l. Comment on the dependence on leg length of the admissible range for stable
walking.
4.15 Numerically simulate the passive walker described in Tutorial 4.1. Use ODE45
in matlab to integrate the equation of motion in Eq. (4.34) with l = 1 m,
α = 60◦, θ (0) = 0◦, and θ̇ (0) = 0.1◦/s. For each stride, you will need one call
to ODE45; the initial conditions of each stride (other than the first one) are given
by Eqs. (4.31) and (4.38). Use Eq. (4.33) to check for toppling, which occurs
if the specific normal force goes to zero. Simulate the walker for 10 steps, or
until it falls over, using ramp angles of 5◦, 12◦, and 18◦. See if you can find
bounds on the ramp angle β for which the walker can move without toppling.
Note that to do this problem, you will need to use the OPTIONS argument of
ODE45 to set two terminal events to stop execution of ODE45. The first event,
which is satisfied at the end of a normal stride, is if the leg angle equals the
value given in Eq. (4.32). The second event, which corresponds to toppling,
is if the specific normal force goes to zero. Look at the documentation for
ODESET for information on creating an event function.
CHAPTER FIVE
Energy of a Particle
In Chapter 4 we integrated Newton’s second law with respect to time to find the change
in linear (or angular) momentum of a particle. We call this result a first integral of the
motion if the momentum is conserved. For impulsive forces, we found the approach
very useful for solving certain problems. Although it does not allow us to compute
entire trajectories (as done in Chapter 3 for some simple systems), it does provide a
convenient formula for finding velocities at different points.
This chapter introduces another integral of the motion—energy—found by integrating Newton’s second law with respect to distance. For certain classes of forces we
can solve for the instantaneous position and/or velocity of a particle without finding
the equations of motion or solving for trajectories. Nevertheless, we still do not introduce any new physics. We simply manipulate Newton’s second law to gain insight
and provide new tools for problem solving.
5.1 Work and Power
Our treatment of the energy of a particle begins with the concept of work, the SI units
of which are joules (kg-m2/s2). The following definition is illustrated in Figure 5.1.
Definition 5.1 Let γP denote the trajectory of a particle P traced by the position
rP/O from rP/O (t1) to rP/O (t2). The work performed on P by force FP along
trajectory γP is
(FP )
WP
(rP/O ; γP ) =
FP . IdrP/O ,
γP
I
vP/O dt, the differential displacement of P relative to O, is
where IdrP/O =
everywhere tangent to γP .
ENERGY OF A PARTICLE
149
I
FP
e2
FP · en
rP/O(t2)
FP · et
γP
P
rP/O(t1)
O
e3
e1
Figure 5.1 Particle P travels along trajectory γP from rP/O (t1) to rP/O (t2 ) under the influence
of force FP . The work done on P by FP along γP is the integral of the tangential component
FP . et from t1 to t2 , where et and en are unit vectors of path frame B = (P , et , en, e3).
Qualitatively, work is the amount of energy it takes to move a particle along a
certain path. The units, joules, are thus units of energy. Anyone who has gone to
the beach knows that it is much easier to drag a wagon full of beach chairs over a
boardwalk than over sand. Dragging the wagon over sand requires more work. This
scenario also depends upon the path. A straight-line path to your desired spot on the
beach involves less work than a circuitous route around various obstacles. Work is an
important tool for finding certain properties of dynamical systems—even though we
integrate out the details of the force.
(F )
The notation WP P (rP/O ; γP ) is clearly a bit cumbersome; nevertheless, it serves
an important purpose. Work is described as a function of the argument rP/O , parametrized by the path γP , since it depends on the path of point P . Even though work
is a scalar quantity, the integral is frame dependent. Because the integration takes
place over changes in the position of P , we specify the frame I as a superscript
on the differential vector displacement IdrP/O (see Appendix A.6). We do not use a
superscript to denote the frame on the work term itself because we assume that work
is calculated in an inertial frame. We label the work with a superscript specifying a
particular force, associating the generation of work with that force.
We can also write the work integral in an alternate form. As long as the particle
doesn’t stop moving along its trajectory, that is, if IvP/O (t) > 0 for all t1 ≤ t ≤ t2,
then the work integral can be written in terms of velocity and time. Simply multiplying
the integrand in Definition 5.1 by dt/dt yields
t2
(F )
WP P (rP/O ; γP ) =
(5.1)
FP . IvP/O dt.
t1
Since the independent variable is now time, we replaced the path integral with a
definite integral between t1 and t2. The specific path is implied by the velocity IvP/O ,
which is the velocity of P along the trajectory γP . This alternate form of the work
integral will prove to be very useful.
It is important to remember that not all forces acting on a particle necessarily do
work. In particular, since the definition of work depends on the dot product of the force
FP with the differential displacement IdrP/O , the component of FP perpendicular to
the trajectory does no work. An important consequence is that a constraint force in a
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CHAPTER FIVE
system with reduced degrees of freedom does no work, as it is always, by definition,
perpendicular to the path. Looking only at the work, then, tells us nothing about
constraint forces.
The alternate form Eq. (5.1) of the work integral naturally leads us to write the
I
vP/O , and the path
work integral in the path coordinate s, defined by ds/dt =
I
I
frame B = (P , et , en, e3), where et = vP/O / vP/O . The work in Eq. (5.1) is then
(F )
WP P (rP/O ; γP ) =
FP . et ds.
γP
We conclude this introduction to the energy of a particle with the definition of
power. While work is a measure of energy, power is a measure of how work changes
over time. Intuitively, power reflects the fact that, over the same period of time, it
takes more work to produce a large force on a particle than it takes to produce a small
force.
Definition 5.2 The power at time t = a of the force FP acting on particle P is the
(F )
rate of change of the work WP P (rP/O ; γP ) evaluated at t = a:
(F )
(FP )
d PP (a) =
WP P (rP/O ; γP ) = FP (a) . IvP/O (a).
dt t=a
The SI units of power are watts (J/s). Note that we used Eq. (5.1) and the second
fundamental theorem of calculus (see Appendix A) to arrive at the second equality in
this definition. To save writing, we often drop the time notation and write
(F )
PP P = FP . IvP/O .
(5.2)
Notationally, power P is distinguished from particle P by the use of Roman type.
Example 5.1 Hoisting a Heavy Mass
Consider the situation illustrated in Figure 5.2a. The objective is to use a rope and
pulley to hoist mass mP to height h. From the free-body diagram in Figure 5.2b we
see that, to lift the mass, the force in the rope T must be greater than or equal to the
weight mP g. Clearly, one criteria of a successful hoist is that the puller be strong
enough to lift the mass. In this example we compute the work done and the power
needed to lift the mass under two different force profiles.
The motion of P is confined to the ey direction. The path γP of the mass is thus
a straight vertical line. From the free-body diagram, Newton’s second law for this
problem is simply
(T − mP g)ey = mP ÿey .
The first approach to hoisting the mass is initially to use the maximum possible
force and then taper off the force until the mass is at the top. This approach is similar
to the clean-and-jerk move used by professional weightlifters. In this scenario, we
ENERGY OF A PARTICLE
151
Tey
I
ey
h
mP
T
–mPgey
mP
ex
(a)
(b)
Figure 5.2 Hoisting mass mP by means of a pulley to height h. (a) Hoist. (b) Reference frame
and free-body diagram.
assume a linear model for the tension force,
y
,
T = mP g + mP b 1 −
h
where b > 0 is a parameter of the model. This force model results in acceleration
y
.
(5.3)
ÿ = b 1 −
h
The force T reaches its maximum at the beginning of the pull when y = 0 and tapers
off to the minimum T = mP g needed to hold the mass at the top of the hoist, when
y = h. The work performed on P by T during the hoist is found from Definition 5.1:
WP(T )(y; γP ) =
h
0
h
T dy =
0
mP
b
= mP h g +
.
2
g+b−
b
y dy
h
(5.4)
It is also interesting to find the power exerted by T during the hoist. To do this
requires the speed of the mass as it rises, which we can then insert into Eq. (5.2). We
find the speed by recognizing that Eq. (5.3) is the equation of motion for the mass,
ÿ + ω02y = b,
√
b/ h. This equation yields simple harmonic motion. Assuming P starts
where ω0 =
from rest, the solution is
y(t) = h 1 − cos(ω0t)
√
ẏ(t) = bh sin(ω0t).
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CHAPTER FIVE
1,500
1,500
1,000
1,000
500
500
Force (N)
Power (W)
0
0.0
0.2
0.4
0.6
y (m)
(a)
0.8
Force (N)
Power (W)
0
0.0
1.0
0.2
0.4
0.6
y (m)
(b)
0.8
1.0
Figure 5.3 The force and power exerted as a 100 kg mass is hoisted to a height of 1 m.
(a) Linear force model with an initial acceleration of 2 m/s2 . (b) Constant-velocity model using
the average velocity from (a).
√
Observe that P reaches
√ y = h at time t = (π/2) h/b, which implies the average
speed is v0 = (2/π) bh. Substituting the tension force T ey , the position yey , and
the speed ẏey into Eq. (5.2) yields
√
PP(T )(t) = mP bh sin(ω0t) g + b cos(ω0t) .
Figure 5.3a shows a plot of the power exerted during the hoist from y = 0 to y = h = 1
m for mP = 100 kg and b = 2 m/s2.
The second approach to lifting the mass is to assume a constant speed. In this case,
ÿ = 0 and, from Newton’s second law, T = mP g during the hoist. In other words, the
puller exerts the minimum force needed to keep the mass moving at constant velocity
v0ey . Note that we are ignoring the initial impulse required to set the mass moving;
we discuss that a bit later.
In this case, the work done by the tension in the rope,
WP(T )(y; γP ) =
h
0
mP gdy = mP gh,
is independent of the speed v0. Furthermore, less work is done in this approach than
in the jerk hoist. The power is also straightforward to calculate. Suppose the constant
√
speed of the mass is given by the average speed during the jerk hoist, v0 = (2/π ) bh.
The power is then
PP(T ) =
2mP g √
bh,
π
which is constant during the hoist. Figure 5.3b shows a plot of the power exerted
during the hoist from y = 0 to y = h = 1 m for mP = 100 kg and b = 2 m/s2.
Which strategy is better? It depends. Certainly, the constant-velocity hoist involves
less total work, which has some appeal. However, it requires a constant exertion of
ENERGY OF A PARTICLE
153
power, which can be tiring. The jerk hoist has a varying power profile, with maximum
power exerted for only a short period of time—interestingly, the maximum power is
not exerted at the same time as the maximum force!
5.2 Total Work and Kinetic Energy
The work integral in Definition 5.1 looks qualitatively similar to the impulse integral in
Definition 4.1 in Chapter 4—in both definitions, force is integrated over an interval.
It is not surprising, then, that we can also find an equation that relates work to the
change of a quantity evaluated at the endpoints of the particle trajectory.
First we need to make explicit the total work WP(tot)(rP/O ; γP ) acting on particle
P . The total work is the work done on P by the vector sum of all of the forces acting
on P over the trajectory γP . This distinction allows us to use Newton’s second law to
replace the force FP in the work integral in Definition 5.1 with the mass, mp , times
the acceleration IaP/O :
WP(tot)(rP/O ; γP ) =
mP
γP
Id dt
vP/O . IdrP/O = mP
I
m
d vP/O . IvP/O = P
2
γP
Id rP/O
d vP/O .
dt
I I
γP
I I
= mP
Iv
P/O (t2 )
Iv
P/O (t1)
d IvP/O . IvP/O .
I
The last integral above contains a perfect differential IdIvP/O 2, which implies that
the integral depends only on the endpoints of the path γP . Integrating results in
WP(tot)(rP/O ; γP ) =
mP
2
Iv
P/O (t2 )
Iv
I
dIvP/O 2
P/O (t1)
m
m
= P IvP/O (t2)2 − P IvP/O (t1)2 .
2
2
=TP/O (t2 )
(5.5)
=TP/O (t1)
We call the quantity TP/O the kinetic energy of P relative to O. We have shown
that the total work on P along γP equals the change in the kinetic energy of P from t1
to t2. We cannot emphasize enough the importance of this result. Even though work,
in general, depends on path, the total work equals the change in kinetic energy.
Definition 5.3 The kinetic energy of particle P relative to O at time t is
1
TP/O (t) =
mP
2
1
vP/O (t) . IvP/O (t) = mP IvP/O (t)2,
2
I
where I is an inertial frame.
Remember that the velocity used in Definition 5.3 must be an inertial velocity. One
of the most common mistakes in dynamics is to use a noninertial velocity in computing
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CHAPTER FIVE
the kinetic energy. For example, this mistake often occurs when computing the kinetic
energy of a particle moving with respect to a noninertial frame (see Chapter 8). Also
note that the kinetic energy depends on the reference point O. The following example
illustrates both of these points.
Example 5.2 Kinetic Energy of a Pendulum
Consider again a pendulum suspended in a box translating relative to inertial frame
I = (O, e1, e2, e3) (see Example 4.6 and Figure 4.8). Here we assume that the box is
traveling at constant velocity, which implies that frame B = (O , b1, b2, b3) is also
inertial. Polar frame C = (O , er , eθ , e3), where er points from O to pendulum bob
P , is not inertial. The kinematics of P relative to O are
rP/O = ler
vP/O = BvP/O = l θ̇eθ ,
I
whereas the kinematics of P relative to O are
rP/O = rO /O + rP/O = xb2 + ler
I
vP/O = ẋb2 + l θ̇eθ .
Using Definition 5.3, we find the kinetic energy of P relative to O is
TP/O =
m
mP I
vP/O 2 = P l 2θ̇ 2,
2
2
which is the typical expression for the kinetic energy of a simple pendulum. The
kinetic energy of P relative to O is
m
mP I
vP/O 2 = P (ẋ + l θ̇ cos θ)b2 + l θ̇ sin θb12
2
2
mP 2 2
=
(l θ̇ + ẋ 2 + 2l θ̇ ẋ cos θ).
2
TP/O =
That is, TP/O and TP/O are different but equally valid measures of the kinetic energy
of P .1 Also, note that using the noninertial velocity CvP/O = 0 to compute TP/O yields
the incorrect result.
Eq. (5.5) leads us to the following work–kinetic-energy formula, which equates the
change in kinetic energy of a particle to the total work done on the particle:
TP/O (t2) = TP/O (t1) + WP(tot)(rP/O ; γP ).
1 Because
(5.6)
the kinetic energy is a quadratic function of the velocity, it does not obey the vector summation
rule analogous to IvP/O = IvP/O + IvO /O . That is, even assuming point O has mass mP , the kinetic
energy TP/O does not equal TP/O + TO /O .
ENERGY OF A PARTICLE
155
What about the relationship between kinetic energy and power? This is easily found
by taking the time derivative of the kinetic energy in Definition 5.3:
d
1
(TP/O ) = mP
dt
2
Id dt
I
vP/O
I . IvP/O + IvP/O . d IvP/O
dt
Id
I
= mP
vP/O .IvP/O .
dt
=FP
I
Using Newton’s second law to replace mP dtd (IvP/O ) with the total force FP acting
on P , we find that
d
(TP/O ) = FP . IvP/O = PP(tot).
dt
By Definition 5.2, FP . IvP/O is the power of FP . To highlight the fact that this
expression is the total power, we once again use the superscript (tot). In words, the
total power acting on P is equal to the rate of change of the kinetic energy of P .
Example 5.3 Carrier Landing
Figure 5.4 shows a fighter jet landing on an aircraft carrier. In this example we use
work to study the forces and acceleration needed to stop the jet before it reaches the
end of the deck. Suppose the jet approaches the aircraft carrier at v0 = 125 knots
(64.3 m/s). Assume that, once the fighter jet touches down on the flight deck, the
arresting cable decelerates it constantly over d = 300 ft (91.44 m) before coming to a
stop. What is the acceleration of the fighter jet and what is the force required to stop
its motion?
Figure 5.4 A fighter jet landing on an aircraft carrier as the tailhook captures the arresting
cable. Image courtesy of Shutterstock.
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CHAPTER FIVE
First we solve this problem using Newton’s second law, assuming that the aircraftcarrier flight deck is an inertial frame. Let e1 be the unit vector in the direction of the
jet’s initial velocity and FP = −F e1 be the (constant) force exerted on the jet by the
carrier’s arresting mechanism. Using the coordinate x for the position rP/O = xe1 of
the jet relative to its position at touchdown, we have
−F e1 = mP ẍe1.
Integrating ẍ =
d ẋ
dx ẋ
= a (constant) over the deceleration,
0
d
ẋd ẋ =
adx,
0
v0
gives
1
− v02 = ad,
2
or
I
1 2
v e1 = −22.6 m/s2e1
2d 0
m
FP = mP IaP/O = − P v02e1.
2d
aP/O = −
Alternatively, using the work–kinetic-energy formula (Eq. (5.6)) and Definition 5.1, we obtain
WP(F )(t1, t2) = TP/O (t2) − TP/O (t1)
d
1
(−F )dx = 0 − mP v02,
2
0
which can be solved to get
FP = −F e1 = −
mP 2
v e1.
2d 0
Using Newton’s second law results in
I
aP/O =
1
1
FP = − v02e1,
2d
mP
as before.
Example 5.4 Take-off from an Aircraft Carrier
This example examines the opposite problem from Example 5.3—an airplane taking
off from the deck of an aircraft carrier. To do so we utilize the expression for power
in Eq. (5.2).
There are two types of planes we want to consider: a propeller-driven aircraft and
a fighter jet. These two types are distinguished by the relationship between thrust
ENERGY OF A PARTICLE
157
and power in the engine. Thrust is the force applied to the airplane by the engine,
causing it to accelerate. A jet engine is characterized by constant thrust at a given
throttle setting. A propeller aircraft is characterized by constant power (determined
by burning fuel in its internal combustion engine). The question is whether enough
power is available in each plane for it to take off before it reaches the end of the carrier
deck.
As in Example 5.3, we take e1 to be the unit vector in the direction of the plane’s
velocity along the deck. The thrust force on the plane is FP = T e1 in the direction of
motion. We first consider the jet. Since the jet applies a constant thrust, the work done
to travel a distance x along the deck (neglecting friction and aerodynamic drag) is
WP(T ) =
0
x
FP . e1dx = T x.
From the work–kinetic-energy formula, the work to accelerate the jet a distance d to
the end of the flight deck is equal to the change in kinetic energy
1
WP(T )(t1, t2) = T d = mP vf2 − 0,
2
where vf is the final speed of the jet when it leaves the deck. We can thus solve for
the final speed generated by the constant thrust T :
vf =
2T d
.
mP
(5.7)
Alternatively, the thrust needed to achieve vf over a distance d is
T =
vf2 mP
2d
.
Using Eqs. (5.2) and (5.7), the power exerted by the jet engine as a function of
distance x is
2T 3x
P=Tv =
.
mP
A typical fighter jet must reach roughly 180 miles/hr in 200 ft to take off from the
deck. The F-14 Tomcat, for example, weighs roughly 61,000 lb fully loaded (27,700
kg) and has a maximum thrust with afterburner of 27,800 lbf (124.7 kN). Plugging
these values into Eq. (5.7) yields the final speed of the jet at the end of 200 ft of only
53 miles/hr, far below the requirement. That is why all aircraft carriers are equipped
with catapults to increase the effective thrust and launch the fighters with the required
speed.
We can perform a similar calculation for propeller-driven aircraft. These are characterized by a constant power P, so the thrust as a function of speed is
T =
P
.
v
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CHAPTER FIVE
The equation of motion for the velocity is given by Newton’s second law,
v̇e1 =
T
P
e1 =
e1.
mP
mP v
Using the same change of variables as in the previous example, v̇ =
can integrate the equation of motion to obtain
d
vf
P
2
v dv =
dx,
mP 0
0
dv dx
dx dt
dv
= v dx
, we
which yields
P
1 3
vf =
d.
m
3
P
Thus the constant power required for the plane to reach the final speed vf in a distance
d is
mP vf3
P=
.
3d
If we consider a plane of comparable weight to the F-14 and ask how much power
is required in the engine to reach 100 miles/hr after 200 ft, rather than 180 miles/hr,
we find that the power needed exceeds 20,000 hp, two orders of magnitude larger than
a typical engine!
5.3 Work Due to an Impulse
What is the work done by an impulsive force? Recall that our model of an impulse is
that the force acts over so short an interval of time that the particle doesn’t move. From
our definition of work, you might reach the conclusion that an impulsive force does
no work. However, the impulse does instantaneously change the inertial velocity of
the particle. By its definition, the kinetic energy is thus instantly changed. Applying
the work–kinetic-energy formula (Eq. (5.6)) contradicts the zero-work conclusion.
The problem here is in our assumption of zero motion during an impulse. This is
only an approximation; in real systems the particle will move, albeit by only a small
amount. When combined with the very large force of an impulse, the result is that
work is in fact done over the short time interval. However, computing the work based
on the force and displacement is not possible, as we don’t have access to the details of
the force or the trajectory. Rather, we use the work–kinetic-energy formula to define
the work done by an impulse.
Definition 5.4 The work WP(FP )(t1, t2) due to an impulse FP (t1, t2) is equal to the
change in kinetic energy before and after the impulse:
(FP )
WP
1
1
TP/O (t2) − TP/O (t1) = mP IvP/O (t2)2 − mP IvP/O (t1)2.
(t1, t2) =
2
2
ENERGY OF A PARTICLE
159
Again, the work done by an impulse is exactly equal to the change in kinetic energy
between t1 and t2. The change in kinetic energy is found by using the velocities before
and after the impulse from the linear-impulse form in Eq. (4.2) or the angular-impulse
form in Eq. (4.17).
Example 5.5 Hoisting a Heavy Mass, Revisited
Recall in Example 5.1 that we brushed aside the impulse required to get the mass
moving for the constant-velocity hoist. In truth, that makes a comparison with the
jerk hoist unfair. Now we treat it more carefully and add in the work to get the mass
moving.
Example 5.1 showed that the work done to lift the mass at constant speed v0 was
independent of the speed and equal to the work against gravity,
WP(T )(y; γP ) =
h
0
mP gdy = mP gh.
The work to lift the mass, however, should include the work to start the mass from
rest. We model the force to initialize the mass at v0 by an impulse; the resulting work
is thus given by the difference in kinetic energy, which in this case is the kinetic energy
of the mass as it rises. The work is
1
WP(T )(y; γP ) = mP gh + mP v02.
2
It is now clear from comparing the work for the constant-velocity hoist to the work
for the jerk hoist in Eq. (5.4) that the constant-velocity hoist involves less work than
the jerk hoist as long as
v0 <
√
bh.
5.4 Conservative Forces and Potential Energy
It is clear from Eq. (5.6) that, if the total work on a particle is zero, then the kinetic
energy of the particle is conserved—this is why kinetic energy is sometimes called
an integral of the motion. Using Definition 5.1, compare Eq. (5.6) to the momentum
integral in Eq. (4.2). Although energy is a scalar and (linear) momentum is a vector,
these two equations are quite similar. Both represent a general solution of Newton’s
second law that, when given a specific force, can be used to find the speed of a particle
at two different points in time.
Recall that, in Chapter 4, we discussed how the momentum integral becomes
extremely useful for certain special classes of forces. This is true for the work–kineticenergy formula as well. There are again three classes of forces for which we make
particular use of the work–kinetic-energy formula:
a. FP is identically zero. That is, the total force acting on particle P is zero. In
this case the total work on P is zero and the kinetic energy is conserved.
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CHAPTER FIVE
I
α
e2
2
β
1
O
e1
e3
Figure 5.5 Two trajectories of a particle from point 1 to point 2. If the particle is acted on by
a conservative force, then the work done along path α is the same as the work done along path
β. If the force is nonconservative, then the work done along α may differ from that along β.
b. FP is constant or is a known function of time. That is, FP is independent of
the position and velocity of the particle. In this (rare) case, the integral in
Definition 5.1 can be solved, either explicitly or numerically, to relate the
kinetic energy at t1 to the kinetic energy at t2.
c. The work done by FP is independent of the path of the particle. We call such
a force a conservative force. In this case—which is the subject of the rest of
this section—Eq. (5.6) becomes particularly useful.
Conservative forces reduce the work integral in Definition 5.1 to a convenient
analytic form. When combined with the kinetic energy in the work–kinetic-energy
formula, we have an extremely useful tool for problem solving and a new conservation law.
5.4.1 Conservative Forces
We begin with a qualitative definition of a conservative force.
Qualitative Definition 5.5 The work performed by a conservative force acting
on a particle depends only on the endpoints of the particle’s path.
The distinction between a conservative force and a nonconservative one is shown
in Figure 5.5. A nonconservative force may perform more (or less) work along path
α than along path β. For a conservative force, the work is the same, since paths α and
β have the same endpoints.
What is the significance of a conservative force? Consider the work done by a force
FP acting on particle P as it first moves on path α from 1 to 2 and then back to 1 along
path −β. The work done while traveling on −β is
(FP )
WP
(rP/O ; −β) =
−β
=−
FP . IdrP/O =
t1
FP . IvP/O dt
t2
t2
t1
(F )
FP . IvP/O dt = −WP P (rP/O ; β),
(5.8)
ENERGY OF A PARTICLE
161
where we have used the alternate form of the work integral from Eq. (5.1). Thus the
work going backward on a path is simply the negative of the work going forward on
the path.
Now suppose the force FP is conservative—that is, the work done by FP is
(F )
(F )
independent of the path. For such a force, WP P (rP/O ; β) = WP P (rP/O ; α) and,
in fact, the work is the same along any path between points 1 and 2. The work done
around the complete loop from 1 to 2 along α and returning from 2 to 1 along −β is
(FP )
WP
(FP )
(rP/O ; α) + WP
(FP )
(rP/O ; −β) = WP
(FP )
(rP/O ; α) − WP
=0=
1
FP . IdrP/O ,
(rP/O ; β)
(5.9)
where 1 denotes the integral around any closed curve through point 1. To obtain
Eq. (5.9), we have used Eq. (5.8) and the fact that the work on the two paths is the
same for a conservative force. Physically, this result means that whatever work is done
by the force in moving away from 1is returned when coming back. We use the fact that
this result holds for any closed curve to introduce the following, more mathematical,
definition of a conservative force.
Definition 5.6 A conservative force is a force FP that satisfies
FP . IdrP/O = 0,
which implies that FP does no work on particle P on any closed trajectory.
The notation WP(c) denotes the work performed on P by the total conservative force,
and WP(nc) denotes the work performed on P by the total nonconservative force. Since
for a conservative force we do not need to indicate the path γP —the work depends
only on the endpoints rP/O (t1) and rP/O (t2)—we write WP(c)(rP/O (t1), rP/O (t2)) or
simply WP(c)(t1, t2). The total work WP(tot)(rP/O ; γP ) is equal to the sum of the work
produced by both conservative and nonconservative forces:
WP(tot)(rP/O ; γP ) = WP(c)(t1, t2) + WP(nc)(rP/O ; γP ).
(5.10)
Example 5.6 A Simple Conservative Force
The simplest conservative force we consider is a constant force field. That is, the force
acting on the particle is independent of its position. A common example is the weight
of a particle near the surface of the earth. The force of gravity is simply FP = −mP gey
everywhere (where the unit vector ey is in the vertical direction). We now show that
this force is conservative using Definition 5.6.
Consider a closed path consisting of the particle rising to height h, moving horizontally a distance l, dropping back down to y = 0, and then moving horizontally
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CHAPTER FIVE
FP
II
I
I
ey
g
III
IV
ex
ey
I
FP
FP
P
P
ex
FP
(a)
(b)
Figure 5.6 Conservative and nonconservative forces. A particle traveling around a closed loop
acted on by (a) a constant gravity force and (b) a constant drag force.
again back to its original position. We label each segment of this square path I–IV,
as shown in Figure 5.6a. The work along segment I is
(F )
WP P (rP/O ; I) = −mP gey . dyey = −mP gh.
I
The work along II is
(F )
WP P (rP/O ; II) =
II
−mP gey . dxex = 0.
The work along III is
(FP )
WP
(rP/O ; III) =
III
−mP gey . dyey = mP gh.
Finally, the work along IV is also zero, since the force is perpendicular to the
trajectory. Summing these together, we find that the work around the closed path
is zero. The constant force thus satisfies Definition 5.6.2
Example 5.7 A Nonconservative Force
This example considers a different force. Let particle P travel the same trajectory
as in Example 5.6 at a constant speed. We find the work done by a drag force
FP = −D Iv̂P/O acting against the velocity with magnitude D = 21 βIvP/O 2, as in
Figure 5.6b. Since the drag force is constant, the work around the closed path is
(F )
WP P (rP/O ; γP ) = −D
h
0
0
dy + D
l
0
dx + D
l
dy − D
h
dx
0
= −Dh − Dl − Dh − Dl = −2D(h + l).
2 You may have noticed that, strictly speaking, we actually did not show that the constant gravity force is
conservative, as we did not show that any and every loop integral was zero, only this specific one. Showing
that every loop integral is zero is quite a bit more complicated; as shown later, we usually determine that
a force is conservative by a different method.
ENERGY OF A PARTICLE
163
B
Nb2
b2
Tb1
T
mP
d
–μcNb1
h
mP
b1
I
–mPgey
θ
(a)
ey
ex
(b)
Figure 5.7 Mechanical advantage. Hoisting mass mP up an inclined plane to height h using
a pulley. (a) Hoist. (b) Reference frames and free-body diagram.
Thus the drag force is not conservative. Because the force is always directed against
the velocity, thus impeding motion, the work done around the closed path is not zero.
This is true for all retarding forces, including viscous damping and Coulomb friction.
When the work done by a force around a closed path is negative, as here, we call the
force dissipative.
Example 5.8 Mechanical Advantage
Consider again the problem of lifting a mass by a pulley (see Example 5.1). Suppose
now, however, that the puller is unable to exert the force necessary to raise the mass
(i.e., Tmax < mP g). In this case, the puller can use mechanical advantage. Rather than
lift the mass directly, he or she places it on an inclined plane and pulls it up the slope,
as in Figure 5.7a. Here we assume a constant-speed lift.
Figure 5.7b shows the reference frames and a free-body diagram for the mass on
the inclined plane. In this problem there are two frames, I and B, both inertial. The
unit vectors of the two frames are related by
ex
b1
b2
cos θ
− sin θ
ey
sin θ
cos θ .
For a constant-speed lift, the acceleration of the mass is zero and Newton’s second
law becomes
T b1 − μc N b1 + N b2 − mP g sin θb1 + mP g cos θb2 = 0,
where N is the normal force and μc is the coefficient of friction.
As a simple case, we begin by ignoring friction (μc = 0), which results in
T = mP g sin θ.
Hence the advantage of using the ramp. The block can be raised to the same height
with a force that decreases with the ramp angle. However, the block must be pulled a
greater distance d = h/sin θ, and thus it will take longer to reach the desired height.
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CHAPTER FIVE
What is the work done by the tension in this case? Following Definition 5.1, we
integrate over the linear path γP using the path coordinate s. We have
WP(T )(s; γP ) =
d
0
T ds = T d = mP gh,
which is independent of θ. By pulling the block up an inclined plane, we gain the
mechanical advantage at seemingly no cost. Because the constant gravity force is
conservative, the work is always mP gh, no matter the path angle.
However, a frictionless surface is rather unrealistic. If instead we include the
friction between the block and the ramp, the tension force becomes
T = mP g sin θ + μc mP g cos θ,
and the resulting work is
WP(T )(s; γP ) =
d
0
T ds = mP gh(1 + μc cot θ).
Thus there is a work penalty if we make the plane angle too small. Since friction is
a nonconservative force, it depends on the path length. For smaller angles, the path
is longer and more work is done to counteract friction. Thus to lift the block with
minimal work, we want to pick the largest possible angle that satisfies the maximum
force limit. Until the angle becomes quite small, the friction cost is very low for
reasonable coefficients of friction.
5.4.2 Potential Energy
Since conservative forces are independent of the path, the line integral in the work
definition can be replaced by a definite integral, which depends only on the endpoints
of the path. We call this integral the potential energy and often write it in indefinite
form.
Definition 5.7 Let FP be a conservative force. The potential energy of a particle
P associated with FP is
−
UP/OP (rP/O ) =
(F )
FP . IdrP/O .
The minus sign in Definition 5.7 is there to be consistent with convention (it makes
later calculations a bit easier). We can always find a potential energy for a conservative
force; Section 5.6 validates this claim.
You may have noticed that, since the potential energy is an indefinite integral, its
value is arbitrary to within an additive constant. Typically, we determine the constant
of integration by choosing a convenient reference point where the potential energy
is zero. Since, as we’ll see next, work is proportional to the difference in potential
energy, any constant chosen simply cancels out.
ENERGY OF A PARTICLE
165
How does the potential energy relate to the work done along a trajectory? Because
the work done by a conservative force is a function only of the endpoints, we can
write it more simply as
(F )
WP P (t1, t2) =
rP/O (t2 )
rP/O (t1)
FP . IdrP/O .
From Definition 5.7 and the rules of definite integration, this work can be written as
(FP )
WP
(F )
(F )
(t1, t2) = UP/OP (rP/O (t1)) − UP/OP (rP/O (t2)).
(5.11)
Thus, the work performed by a conservative force is equal to the negative change
in the potential energy. We rearrange Eq. (5.11) to write the work–potential-energy
formula:
(F )
(F )
(FP )
UP/OP (t2) = UP/OP (t1) − WP
(t1, t2).
(5.12)
Because the potential energy is a (time-varying) function of position, Eq. (5.12)
includes the origin to which the position is referenced (O in Definition 5.7). This
(F )
(F )
notation enables us to write, without loss of clarity, UP/OP (t) or UP/OP to refer to
(F )
UP/OP (rP/O (t)). When it is clear from context, we may also drop the superscript FP .
As with work, later we use the total potential energy UP/O associated with the vector
sum of all conservative forces acting on P . Note that we dropped the superscript (tot)
on the potential energy for brevity; it is understood that UP/O represents the total
potential energy associated with the total force on P .
Example 5.9 The Potential Energy of a Uniform Gravitational Field
When measured close to the surface of the earth, we typically model the gravitational force as a constant with magnitude mP g (where g is the acceleration due to
gravity), rather than use the exact 1/r 2 form of Newton’s universal law of gravity.
Many examples of this appeared earlier in the book. We call this a uniform gravitational field because the acceleration due to gravity is the same everywhere. Here we
compute the potential energy associated with this force model.
Introducing an inertial frame with unit vector ey in the vertical direction, the force
due to the constant gravity field on a particle P of mass mP is simply FP = −mP gey .
Using Definition 5.7, the potential energy is
(F )
UP/OP
=−
(−mP gey ) . dyey = mP g
dy = mP gy,
where we have chosen the constant of integration to be zero so that the potential is
zero at the surface of the earth (y = 0). Note that there is no need to consider the
horizontal displacement of the mass (in the ex direction) as it drops out of the integral
when dotted with the gravity force.
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CHAPTER FIVE
I
x
ey
k
Ney
FP
mP
ex
(a) Coordinates
(b) Free-body diagram
Figure 5.8 Potential energy of a spring-mass system.
Example 5.10 The Potential Energy of a Spring-Mass System
Consider the spring-mass system shown in Figure 5.8a. When the spring stretches or
contracts, it exerts a force FP = −kxex on the mass, where k is the spring constant
and x is the distance the spring is stretched (x is negative for contraction). The spring
force FP is an example of a conservative force. In this example, we compute the
potential energy associated with FP .
The position of the mass is rP/O = xex and its velocity is IvP/O = ẋex . Let x = 0
denote the position of the mass when the spring is not stretched. We use x = 0 as the
reference point for computing the potential energy associated with the spring force.
Using Definition 5.7, we find
1
(F )
UP/OP = − (−kxex ) . dxex = k xdx = kx 2.
2
We have chosen the constant of integration to be zero so that the potential is zero
when the spring is unstretched.
Table 5.1 summarizes some common and important conservative forces and the
corresponding potential energies that arise throughout the book. These forces and
their potentials come up again and again as we solve problems. In each case, we
(F )
derive the potential energy UP/OP using Definition 5.7 and specify the reference point
(F )
at which UP/OP = 0.
An important consequence of Definition 5.7 is that the gradient of the potential energy is a conservative force.(See Appendix A for a review of the gradient operator ∇.)
This result follows from the definition of the directional derivative in Appendix A.6,
Table 5.1
Common conservative forces, their corresponding potential energies, and the
reference points where the potential energies are zero
Force FP
Spring
−kxex
Weight
−mP gey
Gravity
Gm m
− rO2 P er
(F )
Potential energy UP/OP
− (−kxex )dx . ex = 21 kx 2
− (−mP gey ) . dyey = mP gy
Gm m
Gm m
−
− rO2 P er . drer = − Or P
Reference point
x=0
y=0
r =∞
ENERGY OF A PARTICLE
167
(F )
FP = −∇UP/OP .
(5.13)
It is straightforward to verify Eq. (5.13) in Example 5.10. It is an important fact that
every conservative force is the gradient of a scalar potential-energy function. We show
this in the derivations Section 5.6.
It is instructive to verify that replacing the conservative force with the gradient of
a potential leads to the same work–potential-energy formula. Inserting the expression
for FP in Eq. (5.13) into Definition 5.1 and using the fact that FP is conservative
yields
t2
t1
(F )
(F )
(F )
(−∇UP/OP ) . IdrP/O =
∇UP/OP . IdrP/O
WP P (t1, t2) =
t1
t2 (F )
=IdUP/OP
(F )
(F )
= UP/OP (rP/O (t1)) − UP/OP (rP/O (t2)),
where we used the definition of the total derivative from Appendix A.6.
It is natural to ask how we know when a given force is conservative. One way, as
shown in the previous section, is to confirm that the line integral around any closed
path is zero. (It is easy to use this to show that a force is not conservative, but actually
rather tricky to show that every loop integral is zero.) An easier way is to find a scalar
potential-energy function whose gradient is equal to the force. If that can be done,
then the force is guaranteed to be conservative (though this is not necessarily trivial
to do either). Each force in Table 5.1 is conservative because we were able to find a
potential energy for it.
The rate of change of the work associated with a conservative force FP is the power
(F )
(F )
PP P . Just as for kinetic energy, we can show that PP P is equal to the rate of change
of the potential energy:
(F )
PP P = −
d (FP )
U
.
dt P/O
Example 5.11 Computing Work Using Potential Energy
Consider a collar sliding on a shaft connected to a spring, as depicted in Figure 5.9a.
(This system is the subject of Problem 3.20.) The unstretched length of the spring is
l, and the angle between the spring and the shaft is denoted by θ. Since this system
has only one degree of freedom, we need only one coordinate. Let x represent the
coordinate used to describe the position rP/O of collar P on the shaft, where x = 0
at the center of the shaft. Assume that P starts at rP/O = −dex . When the collar is
released, the spring contracts and pulls the collar along the shaft in the ex direction.
We compute the work done by the spring to move the collar from its starting position,
x = −d, at t = t1, to the center, x = 0, at t = t2.
First we compute the work directly from Definition 5.1. Using Figure 5.9b, the
spring force is
FP = kl(cos θex + sin θey ),
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CHAPTER FIVE
O
k
l
I
ey
θ
kΔl(cos θex + sin θey)
P
ex
–Ney
x
(a) Coordinates and reference frame
(b) Free-body diagram
Figure 5.9 Sliding collar attached to a spring.
where l is the amount of stretch in the spring and k is the spring constant. We have
(F )
WP P
=
0
−d
kl(cos θex + sin θey ) . Idxex = k
0
−d
l cos θdx.
(5.14)
Using the geometric relations
l =
x2 + l2 − l
and
x
,
cos θ = − √
2
x + l2
we can integrate Eq. (5.14) to obtain
(FP )
WP
= −k
=k
0
0
−d
−d
x2 + l2 − l √
x−√
lx
x2 + l2
x
x2 + l2
dx
dx
−d
−d 1
1 2
= kx − kl x 2 + l 2 = kd 2 − kl d 2 + l 2 + kl 2.
0
2
2
0
We can also compute the work using Eq. (5.12), which, in this problem, involves
less algebra. As in Example 5.10, the potential energy associated with the spring
force is
1
(F )
UP/OP = kl 2.
2
According to Eq. (5.12), we can compute the work by subtracting the value of the
potential energy at t2 from the value of the potential energy at t1. The initial stretch
in the spring is
l(t1) =
d 2 + l 2 − l.
ENERGY OF A PARTICLE
169
When the mass is at the center, the stretch is zero and thus the potential energy is also
zero. Thus the work done by the spring is
(FP )
WP
(F )
(F )
(t1, t2) = UP/OP (t1) − UP/OP (t2)
1 2
d + l2 − l
= k
2
2
1
− 0 = kd 2 + kl 2 − kl d 2 + l 2,
2
as expected.
You might have guessed that the work is not the same as the change in potential
energy of the spring because the shaft constrains the mass to move in a different
direction than the spring. However, the normal force associated with the constraint
does no work, as with all constraint forces, so the work is due only to the conservative
force from the spring.
5.5 Total Energy
The two work-energy formulas, Eqs. (5.6) and (5.12), relate the work to the change
in kinetic energy of a particle and, for a conservative force, to the change in potential
energy. Eq. (5.10) shows that the total work is also equal to the sum of the conservative
work (work done by conservative forces) plus the nonconservative work (work done
by nonconservative forces). We now show how to combine the work-energy formulas
and eliminate the conservative-work term.
We start by rewriting the work–kinetic-energy formula by separating out the
conservative and nonconservative work:
TP/O (t2) = TP/O (t1) + WP(c)(t1, t2) + WP(nc)(rP/O ; γP ).
Next we replace the conservative work term using Eq. (5.12) and rearrange:
TP/O (t2) + UP/O (t2) = TP/O (t1) + UP/O (t1) + WP(nc)(rP/O ; γP ),
(5.15)
where UP/O (t) is the total potential energy associated with the total conservative force
on P at t. Eq. (5.15) can be written more compactly by defining the total energy.
Definition 5.8 The total energy of particle P at time t is the sum of the kinetic
energy and the potential energy,
EP/O (t) = TP/O (t) + UP/O (t),
where UP/O is the potential energy associated with the total conservative force
acting on P .
Using Eq. (5.15) and Definition 5.8, we obtain our third and most important workenergy formula:
EP/O (t2) = EP/O (t1) + WP(nc)(rP/O ; γP ).
(5.16)
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CHAPTER FIVE
It is worth noting that this formula is why the potential energy is defined as the negative
of the work integral. It allows the total energy to be written as the sum of the kinetic
and potential energies rather than their difference.
By taking the limit of Eq. (5.16) as t = t2 − t1 goes to zero and using Eq. (5.1)
and Definition 5.2, we can write the work-energy formula in differential form as
d
PP(nc).
EP/O (t) = FP(nc) . IvP/O (t) =
dt
In other words, the rate of change of the total energy of P is equal to the power PP(nc)
of the total nonconservative force acting on P .
Eq. (5.16) provides a tool for computing the change in total energy due to nonconservative forces, just as Eq. (4.2) allowed us to compute the change in linear
momentum due to an impulsive force. In the latter case, the conservation of linear
momentum of a particle in the absence of forces is simply a restatement of Newton’s
first law. In the energy case, we have arrived at a new conservation law.
Law 5.1 The law of conservation of total energy of a particle states that, if the total
nonconservative force is zero, then the total energy of a particle is a constant of
the motion.
Law 5.1 follows directly from Eq. (5.16). If every force acting on particle P is
conservative, then WP(nc)(rP/O ; γP ) = 0 and EP/O (t2) = EP/O (t1).
Conservation of energy is an essential tool for solving many dynamics problems. In
fact, it forms the foundation of much of modern physics (as well as thermodynamics).
Without realizing it, we have already used energy conservation to solve problems, as
in Tutorial 3.3. To see how it can be used, we return to our familiar problem, the
simple pendulum.
Example 5.12 Solving the Simple Pendulum Using Energy
This example examines again the simple pendulum, but now using energy concepts.
We use the same kinematics, only now we do not need the acceleration! Let I =
(O, e1, e2, e3), where e1 is oriented downward (in the direction of gravity). In a
rotating frame B = (O, er , eθ , e3) fixed to the pendulum (see, e.g., Figure 3.18), the
kinematics are
rP/O = ler
I
vP/O = l θ̇eθ .
The pendulum is subject to two forces: the force of gravity FP = mP ge1, which is
a conservative force, and the tension in the rod, −F er , which is a normal (constraint)
force. The force of gravity is associated with a gravitational potential energy (see
Table 5.1). Let θ = ± π2 (i.e., when er = ±e2) represent the reference points where
ENERGY OF A PARTICLE
171
the potential energy is zero. The height of the pendulum relative to the reference points
is −rP/O . e1. Using Definition 5.8 and Table 5.1, the total energy of P is
1
EP/O = TP/O + UP/O = mP IvP/O 2 − mP grP/O . e1
2
1
= mP l 2θ̇ 2 − mP gl cos θ.
2
Since the tension in the pendulum rod is a normal force (i.e., it acts in a direction
orthogonal to the velocity of P ), the work done by tension on particle P is zero. We
(T e )
verify this fact by evaluating the work WP r over an arbitrary time interval t1 to t2:
t2
(−T er ) . IdrP/O =
t1
t2
(−T er ) . IvP/O dt =
t1
t2
(−T er ) . ldθeθ = 0.
t1
Since there are no nonconservative forces, WP(tot)(rP/O ; γP ) = WP(c)(t1, t2), and the
total energy remains fixed at the initial value
1
EP/O (0) = mP l 2θ̇(0)2 − mP gl cos θ(0).
2
Using conservation of energy, we can solve for
θ̇ 2 =
2
g
EP/O (0) + 2 cos θ
mP l 2
l
(5.17)
in terms of the initial energy EP/O (0). Note that the result is an equation for θ̇ rather
than θ̈ , which is why the energy formula is an integral of the motion (we have
effectively solved the equation of motion). We came to a similar result in Tutorial 3.3
by directly integrating the equation of motion.
Observe that we obtain the familiar equation of motion for the pendulum simply by differentiating Eq. (5.17). For many single-degree-of-freedom problems, this
approach to finding the equation of motion is less cumbersome than using Newton’s second law because you don’t need to compute the acceleration. However, for
multiple-degree-of-freedom problems, this approach will not work, as there is not
enough information in the (scalar) energy equation to find the equations of motion
for more than one coordinate.3
Of perhaps more interest is the use of the constant total energy to solve for
the trajectory. Solving Eq. (5.17) for θ̇ and integrating with respect to time (using
separation of variables) gives θ as an implicit function of time in the form of the
elliptic integral
t=
θ(t)
θ(0)
dη
2
mP
l2
EP/O (0) + 2 gl cos η
.
(5.18)
3 Chapter 13 discusses an alternate approach to finding equations of motion for any number of degrees of
freedom by starting with the kinetic and potential energies.
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CHAPTER FIVE
It would be great if we could complete this integral to find some function of θ that
could be inverted to find the trajectory θ(t) in terms of the initial angle and initial
energy. Unfortunately, there is no simple closed-form solution for this integral (as
already shown, the pendulum equation of motion has no exact solution in terms
of elementary functions, except for the small-angle case). However, values of this
elliptic integral have been tabulated, and most software packages have built-in elliptic
functions. For instance, this equation is often used to find the period of the simple
pendulum for large initial angles.
5.6 Derivations—Conservative Forces and Potential Energy
In this section, we validate the assertions that (a) for every conservative force, there is
a corresponding potential energy; and (b) if we can find such a scalar potential, then
the force must be conservative.
Theorem 5.1 A force FP associated with a potential energy defined in Definition 5.7
is conservative if and only if it is equal to the negative gradient of the potential
energy:
(F )
FP = −∇UP/OP .
We justify this theorem using Stokes’ theorem. Let be a surface enclosed by a
curve S and n be a unit vector normal to . We have
.
F dr = (∇ × F) . ndA,
S
where dA is a differential area element. Compare the integral on the left-hand side to
the integral in Definition 5.6; if this integral is zero, then F is a conservative force.
(F )
(F )
Assume that UP/OP is a potential energy and FP = −∇UP/OP . It is a property of
the gradient operator that the curl (i.e., ∇×) of the gradient of any scalar function U
is identically zero. (You have the opportunity to show that in the problems.) We have
∇ × ∇U = 0,
which implies
(F )
∇ × FP = −∇ × ∇UP/OP = 0.
Applying Stokes’ theorem, the loop integral of FP is zero, because
FP . dr =
S
Therefore, FP is conservative.
(∇ × FP ) .ndA = 0.
=0
ENERGY OF A PARTICLE
173
Now we validate the “only if” claim. Assume that FP is conservative and define a
(F )
potential energy UP/OP as in Definition 5.7:
(F )
UP/OP (rP/O ) = −
FP . IdrP/O .
Then, from the definition of the directional derivative in Section A.6, we have
(F )
∇UP/OP (rP/O ) = −∇ FP . IdrP/O = −FP ,
which completes the proof.
Theorem 5.1 provides another method to determine whether a force is conservative
—the curl of a conservative force is always zero.
5.7 Tutorials
Tutorial 5.1 Computing the Potential Energy of a Conservative Force
Computing the potential energy of a conservative force is not always as easy as in the
examples in Table 5.1, especially if the force depends on multiple coordinates. This
(F )
tutorial solves for the potential energy UP/OP associated with the conservative force
FP = (−x + y)ex + (x − y + y 2)ey .
(5.19)
Using Cartesian coordinates in reference frame I = (O, ex , ey , ez ), the position of
particle P is rP/O = xex + yey + zez . Using Definition 5.7, we compute the potential
energy (relative to the origin O) as
(F )
UP/OP = − ((−x + y)ex + (x − y + y 2)ey ) . (dxex + dyey + dzez )
=−
(−x + y)dx + (x − y + y 2)dy
=
xdx + (y − y 2)dy − (ydx + xdy)
d(xy)
1
1
1
= x 2 + y 2 − y 3 − xy.
2
2
3
Again, how do we know FP is conservative? Since a potential-energy function
exists, the force must be conservative. We can verify this simply by taking the gradient
(F )
of UP/OP . Alternatively, we can check the curl of the force and verify that it is zero,
since the curl of any conservative force is zero:
∇ × FP = 0.
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CHAPTER FIVE
In Cartesian coordinates using matrix notation, the curl of the position-dependent
force given in Eq. (5.19) is
⎡ ∂ ⎤
⎡
⎡
⎡ ⎤
⎤
⎤
∂
(x − y + y 2)
−x + y
0 − ∂z
0
∂x
⎢ ∂ ⎥
⎢
⎢ ⎥
⎥
⎥
∂
2
⎢
⎥ ×⎢
⎣x−y+y ⎦ =⎣
⎦ =⎣0⎦ ,
∂z (−x + y) − 0
⎣ ∂y ⎦
∂
∂
∂
2
0
0 I
I
∂x (x − y + y ) − ∂y (−x + y) I
∂z I
which implies that FP is indeed conservative.
Tutorial 5.2 Energy in Orbital Motion
Now we look again at the orbit of a particle under a central force FP . Chapter 4 showed
that the equations of motion for the orbit of a particle under gravitational attraction is a
conic section whose eccentricity (a constant of integration) determines the shape of the
orbit. The approach taken in Chapter 4 used conservation of the angular momentum
of the particle.
Because gravity is a conservative force, the total energy of an orbit should also be
a constant of the motion (as there are no nonconservative forces acting). As usual, the
energy is
1
EP/O = mP IvP/O 2 + UP/O .
2
Using IvP/O = ṙer + r θ̇eθ and UP/O = −GmO mP /r (see Table 5.1), the specific
energy (i.e., energy per unit mass) is
εP =
1
1
μ
EP/O = (ṙ 2 + r 2θ̇ 2) − ,
mP
2
r
where μ =
GmO .
2
r θ̇ of the orbit
Recall from Tutorial 4.2 that the specific angular momentum hO =
is constant. This allows us to write the energy entirely in terms of r:
h2
1
μ
εP = ṙ 2 + O2 − .
2
2r
r
(5.20)
This equation looks like the energy of a single-degree-of-freedom system with coordinate r, kinetic energy 21 ṙ 2, and an effective potential
Ueff (r) =
h2O
2r 2
−
μ
.
r
The concept of the effective potential arises frequently in dynamics. It often allows
us to reduce a complicated system to what looks like a single degree of freedom.
For instance, for a given set of initial conditions, we could compute the energy and
angular momentum and then find ṙ(t) at any r(t) from Eq. (5.20).
The effective potential also determines the turning points of the orbit, that is, the
points where ṙ goes to zero and thus the trajectory turns around. The trajectory must
turn around at such points, since the kinetic-energy–like term in Eq. (5.20) is always
positive; the effective potential is a maximum when ṙ = 0. (This is another way of
ENERGY OF A PARTICLE
175
showing that ṙ = 0 at periapsis and apoapsis.) We can write the energy in terms of rp ,
the periapsis radius:
εP = Ueff (rp ) =
h2O
2rp2
−
μ
.
rp
Using the polar equation for the periapsis distance in Eq. (4.53), we can also write
the energy in terms of the angular momentum and eccentricity,
εP = −
μ2
(1 − e2).
2h2O
(5.21)
Note this implies εP < 0 when e < 1 (circular and elliptical orbits).
This result can be used to find a new expression for the eccentricity of the orbit in
terms of the specific angular momentum and energy:
2εP h2O
.
(5.22)
e = 1+
μ2
Thus the angular momentum and energy can be used as constants of the trajectory
in lieu of the initial conditions. Eq. (5.22) shows how the eccentricity of the orbit is
determined by the specific angular momentum and the specific energy.
Finally, we can substitute from Eq. (4.55) for the semimajor axis into Eq. (5.21)
to find
μ
,
a=−
2εP
which implies that the semimajor axis a of an elliptical orbit is entirely determined by
the specific energy, εP ! The new expressions for a and e show that the two constants
of the motion, εP and hO , determine the size and shape of the orbit.
Tutorial 5.3 Energy of a Two-Degree-of-Freedom System
Up to now, every example we have examined using energy ideas was a single-degreeof-freedom system. Energy is particularly useful for such systems because the workenergy formula can be used to determine the state of the system at some time given
its state at an earlier time (such as the speed of a falling ball when it hits the ground).
We could even obtain the equation of motion by differentiating the total energy.
Yet, many (actually, most) systems have more than one degree of freedom and are
thus described by multiple coordinates. But we have only one formula for conservation
of energy (a scalar equation). That does not provide enough information to completely
determine the state of the system. We can also no longer find the equations of motion
by simply differentiating the work-energy formula. Is energy only a useful concept
for single-degree-of-freedom problems? How might it help for this broader class of
problems?
Conservation of energy is still an extremely helpful property and something you
should always examine. As problems become more complicated, keeping track of
energy and verifying that it remains conserved provides a helpful check that you are
staying on course. We often use it in numerical integration to check the performance
of the code. It also can provide useful information on the global behavior of a system,
176
CHAPTER FIVE
I
ey
B
eθ
–k(r – r0)er
O
g
ex
r
er
θ
mP
P
P
(a) Coordinates
(b) Reference frames
–mPgey
(c) Free-body diagram
Figure 5.10 Pendulum attached to a spring.
even when specific trajectories might be very complex. We show such a case in this
tutorial, where we introduce the concept of a zero-velocity curve.
Problem 3.23 asked you to solve for the equations of motion for the simple spring
pendulum. Figure 5.10 shows this system again. This is a two-degree-of-freedom
problem best solved using polar coordinates. Unlike the simple pendulum, where the
radial position is a fixed length, here r varies along with θ. The position and velocity
of the pendulum are thus given by the usual polar coordinate equations,
rP/O = rer
I
vP/O = ṙer + r θ̇ eθ .
We can use Newton’s second law in polar coordinates to find the equations of motion,
r̈ = r θ̇ 2 + g cos θ −
θ̈ = −
k
(r − r0)
mP
2ṙ θ̇
g sin θ
−
,
r
r
where r0 is the unstretched length of the spring. These equations are quite complicated
and have no exact solution. We show a particular simulated trajectory in Figure 5.11.
Looking at the total energy gives us information about all possible trajectories of
P . The kinetic energy of P is found from the velocity:
1
1
TP/O = mP IvP/O . IvP/O = mP (ṙ 2 + r 2θ̇ 2).
2
2
The potential energy of the particle arises from the spring and from gravity,
1
UP/O = −mP gr cos θ + k(r − r0)2.
2
The total energy is thus
1
1
EP/O = mP (ṙ 2 + r 2θ̇ 2) − mP gr cos θ + k(r − r0)2.
2
2
(5.23)
4.5
4.0
3.5
3.0
2.5
2.0
1.5
1.0
0.5
0.0
177
2.0
1.5
1.0
θ (rad)
r (m)
ENERGY OF A PARTICLE
0.5
0.0
–0.5
–1.0
–1.5
0
5
10
15
t (s)
20
25
30
–2.0
0
5
10
(a)
15
t (s)
20
25
30
(b)
Figure 5.11 The r and θ trajectories for a simple spring pendulum of unstretched length
r0 = 0.5 m, spring constant k = 6 N/m, and initial conditions r(0) = 0.01m and θ(0) = π/3 rad.
(a) Radial position. (b) Angle of pendulum.
The trajectories of P must be such that, given the initial conditions, EP/O is a
constant. We can say more, however. Observe from Eq. (5.23) that the kinetic energy
term is always positive or zero; it is quadratic in the rates of both coordinates, ṙ and θ̇.
Thus for any point on the trajectory where ṙ = 0 and θ̇ = 0, the energy is a minimum.
We plot potential energy contours as a function of (r, θ )I and know that, whatever
the trajectory, it must always be on one side of the potential contour equal to the
initial total energy. These lines are called zero-velocity curves. They represent the
configuration of the system at zero velocity.
For example, Figure 5.12 shows the zero-velocity curves for the simple spring
pendulum using the same parameters as in Figure 5.11. We also overlay a plot of
(r, θ )I for the trajectory from Figure 5.11. When the two coordinate rates go to zero,
the trajectory touches the zero-velocity curve. The touch points are called turning
points because the trajectory must turn around at these points, since ṙ and θ̇ both go to
zero. Plots such as these yield a great deal of information about where the trajectories
must reside and what form they take without our actually solving the equations of
motion.
Contour plots such as Figure 5.12 are particularly useful for nonlinear systems
that exhibit chaotic or almost chaotic behavior. These systems are characterized by
extreme sensitivity to initial conditions. In other words, changing the initial conditions
by a small amount results in a dramatically different trajectory. For instance, for the
simple spring pendulum simulated here, a small change in the initial angle produces
a trajectory indistinguishable from the one in Figure 5.11. However, if we modify the
system slightly by making the spring nonlinear such that the spring force is
Fp = −k1(r − r0)er − k2(r − r0)3er ,
very small changes in the initial angle result in significantly different trajectories. For
example, Figure 5.13 shows two such trajectories for r(0) = 0.46 m and two different
initial angles, θ (0) = 1.05 rad and θ(0) = 1.06 rad. Even though the resulting motion
178
CHAPTER FIVE
8
7
6
r (m)
5
4
3
2
1
–3
–2
–1
0
θ (rad)
1
2
3
Figure 5.12 Zero-velocity curves (solid lines) associated with a simple spring pendulum
with parameters r0 = 0.5 m and k = 6 N/m. Dashed lines are the trajectories of the system in
Figure 5.11.
1.5
2.5
1.0
2.0
θ (rad)
r (m)
0.5
1.5
1.0
–0.5
0.5
0.0
0.0
–1.0
0
5
10
15
t (s)
(a)
20
25
30
–1.5
0
5
10
15
t (s)
20
25
30
(b)
Figure 5.13 The r and θ trajectories for a nonlinear spring pendulum of unstretched length
r0 = 0.305 m, linear spring constant k1 = 6 N/m, nonlinear spring constant k2 = 3 N/m3, and
initial conditions r(0) = 0.46 m, θ (0) = 1.05 rad (solid curves), and θ(0) = 1.06 rad (dashed
curves). (a) Radial position. (b) Angle of pendulum.
is quite different, the total energy for the two sets of trajectories is very close. The
zero-velocity curves thus provide useful information on the behavior of families of
trajectories without having to integrate specific ones. Figure 5.14 shows the constantenergy contours for the nonlinear-spring system overlaid with the two trajectories
shown in Figure 5.13.
ENERGY OF A PARTICLE
179
4.0
3.5
3.0
r (m)
2.5
2.0
1.5
1.0
0.5
0.0
–3
–2
–1
0
θ (rad)
1
2
3
Figure 5.14 Zero-velocity curves (solid lines) associated with a nonlinear spring pendulum
with parameters r0 = 0.305 m, k1 = 6 N/m, and k2 = 3 N/m3. Dashed lines are the trajectories
of the system in Figure 5.13.
5.8 Key Ideas
.
The work performed on particle P by force FP along trajectory γP is
(FP )
WP
(rP/O ; γP ) =
FP . IdrP/O .
γP
The work performed on P by the total force acting on P is denoted
WP(tot)(rP/O ; γP ).
.
The power exerted by force FP acting on P at time t is the instantaneous rate of
change of the work,
FP (t) . IvP/O (t).
PP P (t) =
(F )
.
The kinetic energy of particle P at time t is
1
mP
TP/O (t) =
2
1
vP/O (t) . IvP/O (t) = mP IvP/O (t)2.
2
I
The change in kinetic energy from t1 to t2 equals the total work on P along
trajectory γP from rP/O (t1) to rP/O (t2):
TP/O (t2) = TP/O (t1) + WP(tot)(rP/O ; γP ).
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CHAPTER FIVE
.
The work performed by impulse FP (t1, t2) is equal to the change in kinetic energy
across the impulse,
(FP )
WP
.
.
(t1, t2) = TP/O (t2) − TP/O (t1).
A conservative force does no work on any closed trajectory:
FP . IdrP/O = 0.
The work generated by a conservative force on particle P depends only on the
endpoints of γP . We decompose the total work into the sum of the conservative
work and the nonconservative work:
WP(tot)(rP/O ; γP ) = WP(c)(t1, t2) + WP(nc)(rP/O ; γP ).
.
The potential energy of particle P associated with conservative force FP is
(F )
UP/OP (rP/O ) =
− FP . IdrP/O .
A conservative force can be represented as the gradient of a potential energy,
(F )
FP(c) = −∇UP/OP ,
and the negative change in the potential energy from t1 to t2 equals the conservative
work,
UP/O (rP/O (t2)) = UP/O (rP/O (t1)) − WP(c)(t1, t2).
.
The total energy of particle P at time t is the sum of the kinetic energy and the
potential energy
EP/O (t) = TP/O (t) + UP/O (t).
.
The work-energy formula,
EP/O (t2) = EP/O (t1) + WP(nc)(rP/O ; γP ),
implies that total energy is conserved if the total nonconservative force acting on
P is zero.
5.9 Notes and Further Reading
The material in this chapter is exceptionally important and forms the foundation
of much of modern physics. In particular, an understanding of conservative forces
and potential energy is critical. In addition to providing important tools for solving
problems, they are the foundation of your later courses in dynamics. In fact, there
is a fascinating historical debate over what is more fundamental, forces or energy.
ENERGY OF A PARTICLE
181
As noted in Chapter 2, the entire concept of force is problematic. Most modern
approaches to mechanics (and quantum mechanics) begin with kinetic and potential
energy as the fundamental physical quantities. We introduce you to this approach in
Chapter 13. The historical implications are discussed in the article by Wilczek (2004).
By now it is clear that we have a particular fondness for the simple pendulum, both
for the elegance of the problem and its pedagogical value. In fact, you probably have
noticed that the simple pendulum is not so simple (there is a lovely article with that
name that discusses the complexity of the pendulum by Antman [1998]). This chapter
introduced a solution approach to the pendulum using energy. For readers interested in
delving further into the large-angle solution of the pendulum, Greenwood (1988) has
an excellent and concise discussion involving elliptic functions. The pendulum is also
often used as the foundation for studying and teaching chaos theory. An introductory,
though mathematically advanced, book on chaos is the one by Guckenheimer and
Holmes (2002).
5.10 Problems
5.1
You are driving a car down a road at constant speed v0 when you see a deer
in your headlights. After a reaction time t0 you slam on your brakes and your
wheels lock. There is a constant friction force Ff between your wheels and the
road. Ignoring air drag, find how far the car will travel after you see the deer
before coming to a stop. Assume that t0 = 0.75 s, v0 = 45 mph, and μk = 0.7.
State any additional assumptions you make while solving the problem.
5.2
Consider a child on a sled going down a hill of height 5 m, as depicted in
Figure 5.15. You can assume the hill is frictionless. The child and sled have
a mass m of 20 kg. Once on the ground, the friction brings the sled to a stop
within 10 m. What is the coefficient of friction?
m
5m
10 m
Figure 5.15 Problem 5.2.
5.3
In the log flume you are dropped from height h, and at the very bottom you
are turned horizontal and brought to a stop by the water. Assume that you and
the log have mass m. Is the system conservative? What is the work done by
the water to bring you to a stop?
5.4
Consider again Problem 3.20. If the collar starts at a distance x0 from the
center C of the shaft, what is its speed when it passes through C?
182
CHAPTER FIVE
5.5
Show that the potential energy of the gravity force from Newton’s universal
law of gravity, −Gm0mP /r 2, is correct as given in Table 5.1.
5.6
Recall Problem 2.13. Is such a nonlinear-spring force conservative? If so, then
what is the potential-energy function? If not, then explain why not.
5.7
Consider the attracting dipole of Problem 4.7. Using the expression for the
gravitational potential of a particle, show that the potential due to the dipole
at a position (r, θ)I is
2
mP μ
d
(FP )
2
UP/O (r, θ) ≈ −
3 cos θ − 1 ,
2+
r
r
where μ = Gm and P is far from the dipole (r d). Retain terms only to the
first and second powers in (d/r). Show also that this potential corresponds to
the force given in Problem 4.7. Finally, is the energy of the resulting motion
in Problem 4.7 conserved?
5.8
Consider a mass-spring system with mass m of 1.5 kg and spring constant
k of 0.2 N/m, as shown in Figure 5.16. The spring rest length is 0.5 m. Suppose
the mass is initially located at x = 0.2 m. What is the potential energy of the
mass due to the spring force? If the mass is released from rest, what will its
velocity be when it reaches x = 0.6 m?
ey
k
m
ex
Figure 5.16 Problem 5.8.
5.9
Mass m is projected up a fixed ramp with initial speed v0, as shown in Figure 5.17. The ramp angle is θ and its coefficient of friction is μ. What is the
maximum height that the mass will reach?
g
v0
m
θ
Figure 5.17 Problem 5.9.
5.10 Mass m on a string of length l is released from rest at θ = 30◦, as shown
in Figure 5.18. The string encounters a fixed obstacle that is half as long as
the string, but the mass keeps swinging. What is the angle that the mass will
reach at the leftmost point of its swing?
ENERGY OF A PARTICLE
183
O
O
l
–
2
g
θ
l
m
Figure 5.18 Problem 5.10: Initial (black) and final (white) mass positions.
5.11 Suppose satellite P is in a circular orbit of radius ρ around a large fixed mass
O, as shown in Figure 5.19. How much energy is required to change to an
elliptical orbit of eccentricity e and semimajor axis a? [HINT: See Tutorial
5.2.]
O
P
ρ
2a
Figure 5.19 Problem 5.11.
5.12 Mass m slides in a frictionless parabolic bowl defined by y = x 2, where (x,y)I
are Cartesian coordinates in an inertial reference frame I = (O, ex , ey , ez ),
as shown in Figure 5.20.
a. Draw a free-body diagram for mass m. (There is gravity in this problem.)
b. Using only the Cartesian coordinate x, find the position rP/O and inertial velocity IvP/O of the mass with respect to O. [HINT: Remember
y = x 2.]
c. Find the potential energy UP/O of the mass with respect to O.
d. Find the kinetic energy TP/O of the mass with respect to O.
e. Show that the total energy of the mass with respect to O is conserved.
I
ey
m
g
P
y = x2
O
ex
Figure 5.20 Problem 5.12.
184
CHAPTER FIVE
5.13 Figure 5.21 shows two tracks connecting points A and B. A car can be started
from point A with zero initial velocity and will follow either track to point
B, propelled only by gravity (friction should be taken as negligible). The first
track connects the points in a straight line (making it the shortest-distance
route) and has length l and a vertical drop of h. The second track is a very
special shape called a cycloid and is the shortest-time route between points A
and B. The shortest-time trajectory is the solution to the brachistochrone problem (from Greek—brachistos, meaning “the shortest,” and chronos, meaning
“time”), which was one of the earliest problems solved using the calculus of
variations.
A
l
h
g
B
Figure 5.21 Problem 5.13.
a. For the straight track, find the time it would take the car to travel from
point A to point B and its velocity at B. Do so using Newton’s laws
and then by using conservation of energy. Ensure that the results from
your two methods match. All calculations for this part should be done
analytically, without the use of matlab or any other computer tools.
b. Write out the total energyof a car
on a cycloid track in terms
traveling
v. If the car starts at rest with zero
of its height y and speed IvP/O =
height, we can define the total energy of the car to be zero. Show, then,
that the speed of the car as a function of height can be written as
v = −2gy.
c. Find the velocity at point B of a car traveling on the cycloid track.
5.14 The cycloid discussed in Problem 5.13 can be represented parametrically as
C
(φ − sin φ)
2
C
y(φ) = − (1 − cos φ),
2
x(φ) =
where φ is the parametric parameter (varying between 0 and 2π), and C is a
constant of the cycloid, determined by the endpoints.
a. We can express the velocity of a car traveling on the cycloid track as
v = ds
dt because the velocity is constrained to be along the path. We
can also write ds 2 = dx 2 + dy 2. Equate the two expressions for ds,
ds = vdt = dx 2 + dy 2,
ENERGY OF A PARTICLE
185
and divide both sides by dφ to get
dx 2 + dy 2
dt
v
=
.
dφ
dφ
dt
and integrate.
Because x(φ) and y(φ) are known, we are able to find dφ
√
Using v = −2gy, show that the relationship between φ and t is
C
t=
φ.
2g
b. If the length of the straight track l is 1 m and its vertical drop h is 12
cm, the constant of the cycloid, C, will equal −0.58 and φ will equal
1.59π at point B (assuming that it is 0 at point A). Using these values,
find how much time it would take a car to travel from A to B when
following either the straight track or the cycloid track in Figure 5.21.
c. Using matlab, numerically integrate the equations of motion for a
car following the cycloid track and verify the time you calculated in
part (b) for the car to reach point B.
d. Using matlab, see if you can find the values of C and φ at point B
given in part (a). You may find the function fsolve useful here, but
there are many different ways of doing this.
5.15 Show that the curl of the gradient of any scalar function is identically zero.
5.16 Consider a simple circular loop-the-loop. The car enters the circular loop of
radius Rc with horizontal speed v0.
a. Find an expression for the speed of the car as a function of the angle
around the loop.
b. Find the inertial acceleration of the car as a function of v0 and θ and
express it in a frame fixed to the car.
c. What is the normal force applied to the car by the track?
d. What is the minimum speed v0 needed to complete the loop?
5.17 Consider the pile driver shown in Figure 5.22. Mass M of 50 kg is dropped
on top of the pile, where it stays. The pile (mass m = 10 kg) is stationary until
M
2m
m
Figure 5.22 Problem 5.17.
186
CHAPTER FIVE
M hits it. The large mass is dropped from a height of 2 m. After the impact
the pile is driven into the ground by an additional 5 cm. What is the force of
resistance of the ground to the combined pile/mass motion?
5.18 A block of mass mP is initially sitting at rest at the bottom of a frictionless,
circular loop of radius R, as shown in Figure 5.23. Gravity acts downward.
An impulse to the right is then imparted to the block.
∗
a. What minimum impulse FP must be imparted to the block to ensure
that it can make it all the way around the hoop without losing contact?
∗
b. If an impulse of FP = √2 FP is applied to the block (still to the right),
5
at what angle θ (measured counterclockwise from the downward direction) does the block lose contact with the hoop?
R
g
θ
mP
Figure 5.23 Problem 5.18.
5.19 In Problem 5.18, at what angle θ does the block strike the hoop again if given
the impulse in part (b)?
5.20 Consider particle P of mass mP starting at height h and attached to a slack
string of length 2h/3 a distance h/3 away. The mass falls vertically downward
under gravity. When the string goes taut, the mass will start to swing (assume
that the string does not stretch).
a. What is the linear impulse imparted to the mass at the moment the
string pulls taut?
b. How high will P rise on the other side as it swings as a pendulum?
5.21 Suppose you are driving your new SUV that gets 25 mpg highway at 55 mph
(the speed for which the mpg was calculated). You drive from Baltimore to
Philadelphia and back, roughly 140 miles. First calculate how many gallons
of gas you use. Then estimate how many gallons you burn if you drive
to Philadelphia and back at 70 mph and at 80 mph. Be sure to state your
assumptions and modeling steps. Assume the road is flat for the entire drive
and ignore any frictional losses.
PART TWO
Planar Motion of a Multiparticle System
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CHAPTER SIX
Linear Momentum of a Multiparticle System
Up to now we have focused on the dynamics of one particle only. In fact, we made
a particular effort in Chapter 2 to point out that Newton’s second law applies only
to a single point mass. What to do, then, when faced with many particles? This
chapter answers that question. We examine dynamics problems involving pairs of
and sometimes many interacting particles. Does that mean there is a need for new
physical concepts? In short, no. Treating multiple particles simply entails using
Newton’s second law on each particle individually. The result, of course, is that the
number of degrees of freedom increases with the number of particles, as does the
corresponding number of equations of motion. Nevertheless, studying the dynamics
of many particles is no more complicated than solving Newton’s second law (or its
integrated form) for each particle. However, as the number of particles increases, this
task can quickly become unwieldy. We therefore introduce some important concepts
concerning the linear momentum of multiparticle systems that enhance understanding
and simplify certain problems. These ideas are essential to the developments in the
remainder of the book.
6.1 Linear Momentum of a System of Particles
We begin by generalizing to many particles the study of linear momentum and
Newton’s laws from Chapters 3 and 4. As before, we use Newton’s second law,
keeping track of the degrees of freedom and coordinates for each particle. The main
addition is that we carefully consider the forces among the particles (called internal
forces). This concept is examined in the first subsection below.
For some multiparticle systems we gain understanding and/or reduce complexity
by studying the linear momentum of the entire collection of particles as a whole.
The resulting tools sometimes dramatically simplify problems and prove useful when
attacking more complicated systems. Sections 6.1.2 and 6.1.3 examine these ideas.
190
CHAPTER SIX
I
e2
i
Fi,j
Fj,i
j
O
Fj(ext)
e1
e3
Figure 6.1 Internal and external forces acting on a collection of many particles.
6.1.1 Newton’s Second Law for Multiple Particles
There are several notational conveniences adopted here that will simplify the treatment. In most problems with more than one particle, we separate the forces acting
on the particles into external and internal forces (see Figure 6.1). External forces are
field or contact forces coming from outside the system that act on one or more of the
particles independently of the other particles. Internal forces are those that act among
the various particles (e.g., gravitational or electrostatic attraction), usually along the
lines between particles. The notation Fj,i indicates the force acting on particle j due
to particle i.1 Fj(ext) is the total external force acting on particle j .
Using multiparticle notation, Newton’s second law for particle j is
I
mj aj/O = Fj(ext)
N
+
Fj,i .
(6.1)
i=1
We make explicit the assumption that a particle cannot exert a force on itself by setting
Fj,j ≡ 0.
To solve many multiparticle problems, Eq. (6.1) is all that is needed. We simply
solve a system of N vector equations.
Example 6.1 The N-Body Problem
One of the most famous (and earliest) multiparticle dynamics problems is the collective motion of the sun and planets under their mutual gravitational attraction. The
general problem of finding the equations of motion for N gravitationally bound particles is known as the N -body problem. This problem is illustrated in Figure 6.1 with
Fj(ext) = 0 and the internal forces given by Newton’s universal law of gravitation,
1 We
often label particles using lowercase letters, such as i and j , which represent a number from one to
N , where N denotes the total number of particles.
LINEAR MOMENTUM OF A MULTIPARTICLE SYSTEM
Fj,i = −
Gmj mi
ri/O − rj/O 3
191
rj/i ,
where rj/i = rj/O − ri/O . The equations of motion for each particle (planet or star)
are thus given by Eq. (6.1):
Id 2 dt 2
rj/O +
N
i=1,i =j
Gmi
rj/i = 0,
ri/O − rj/O 3
j = 1, . . . , N.
This is a system of N coupled, vector differential equations that must be solved simultaneously. Unfortunately, unlike the case studied earlier of a single body attracted
toward a fixed origin, there is no solution of this problem in terms of elementary
functions. The best we can do for this multiparticle problem is to write down the separate equations of motion and study them numerically (or analytically under certain
approximating assumptions).
In some cases we can reduce the number of scalar equations of motion by recognizing explicit constraints that reduce the number of degrees of freedom. For example, in
this chapter, as before, we assume that each particle is constrained to move in a plane
only. In some problems, the particles are rigidly connected (forming a rigid body,
discussed in detail later) or semirigidly connected (e.g., by means of a connection
that allows motion in only one direction, like a linear spring). A common approach
for these problems is to “break” the connection, introduce internal constraint forces
(as often done with single-particle problems), write the equations of motion for each
particle, and then eliminate the internal constraint forces as before. The next example
illustrates this procedure.
Example 6.2 The Crane
Consider a simple pendulum P hanging from block Q, which is free to slide horizontally (Figure 6.2). There are two degrees of freedom in this problem: the horizontal
motion of the block and the swing of the pendulum. We thus use the scalar coordinates x and θ to describe, respectively, the position of the crane and the angle of the
pendulum relative to the vertical (Figure 6.2a). In this problem we find the equations
of motion for x and θ by breaking apart the block and pendulum and accounting for
the internal tension in the rod.
Let I = (O, ex , ey , ez ) denote the inertial frame and B = (Q, er , eθ , ez ) denote a
polar frame fixed to the pendulum. The frames B and I are related by the transformation
ex
ey
er
eθ
sin θ
cos θ
− cos θ
sin θ .
We start by writing the kinematics of the two masses P and Q:
rQ/O = xex
I
vQ/O = ẋex
I
aQ/O = ẍex ,
(6.2)
192
CHAPTER SIX
I
ey
g
x
eθ
Q
O
O
Q
ex
G
θ
Ney
B
Q
Ter
–mQgey
l
–Ter
P
er
P
P
–mPgey
(a) Coordinates
(b) Reference frames
(c) Free-body diagrams
Figure 6.2 A crane. G is defined in Example 6.6.
and, using rP/O = rQ/O + rP/Q,
rP/O = xex + ler
I
vP/O = ẋex + l θ̇eθ
I
aP/O = ẍex + l θ̈eθ − l θ̇ 2er .
Using the free-body diagram Figure 6.2c and Newton’s second law on each mass
gives
(N − mQg)ey + T er = mQẍex
−mP gey − T er = mP (ẍex + l θ̈eθ − l θ̇ 2er ).
(6.3)
(6.4)
Using the transformation in Eq. (6.2), the two vector expressions in Eqs. (6.3) and (6.4)
yield four scalar equations relating the unknown quantities ẍ, θ̈, N , and T . These four
equations can be solved algebraically for the equations of motion in x and θ:
mP
m
m
2
1−
cos θ ẍ − P g sin 2θ − P l θ̇ 2 sin θ = 0
2mG
mG
mG
mP
m
g
2
1−
cos θ θ̈ + P θ̇ 2 sin 2θ + sin θ = 0,
l
2mG
mG
(6.5)
(6.6)
where mG = mP + mQ.
These equations can be integrated to find the motion of the pendulum mass and
the block. They can also be used to design a controller for an overhead crane.
6.1.2 Total Linear Momentum and Momentum Conservation
This subsection develops an idea that can be used to study the motion of a collection
of particles without necessarily resorting to simultaneously solving the equations of
LINEAR MOMENTUM OF A MULTIPARTICLE SYSTEM
193
motion for each particle. We begin by integrating Eq. (6.1) to obtain the impulse form
of Newton’s second law, Eq. (4.2), for each particle:
(ext)
mj Ivj/O (t2) = mj Ivj/O (t1) + Fj
N
(t1, t2) +
Fj,i (t1, t2),
(6.7)
i=1
t2
where Fj,i (t1, t2) = t Fj,i dt is the linear impulse acting on particle j from t1 to t2
1
due to the internal force of particle i. Again, there is nothing new here, other than being
careful to indicate which particle we are considering. There are N different (vector)
equations for the change in momentum of each particle. However, what happens if
we sum over all particles? This leads us to define a new quantity, the total linear
momentum.
Definition 6.1 The total linear momentum IpO of a collection of particles is the
vector sum of the individual linear momenta:
N
I
pO =
mj Ivj/O .
(6.8)
j =1
Next we use Definition 6.1 to find the integrated form of Newton’s second law for
the total linear momentum by summing over all particles in Eq. (6.7):
pO (t2) = IpO (t1) +
I
N
(ext)
Fj
N
N
(t1, t2) +
Fj,i (t1, t2).
(6.9)
j =1 i=1
j =1
Now we use Newton’s third law of motion, which states that Fi,j = −Fj,i . The
!N
!
implication of this (and the assumption Fj,j ≡ 0) is that N
j =1
i=1 Fj,i (t1, t2 ) = 0.
(To see this, imagine creating an N × N matrix with the (i, j )th entry equal to Fi,j
for all pairs i and j . Summing over the rows and columns of the matrix yields zero
because the diagonal entries are zero and all of the off-diagonal entries cancel out.)
Thus, Eq. (6.9) becomes
I
pO (t2) = IpO (t1) + F
(ext)
,
(6.10)
where F(ext) is the total external force acting on the system. This result is extremely
important; the total linear momentum of the collection of particles obeys Newton’s
second law when acted on by an external force. The internal forces completely cancel
out.
Eq. (6.10) also leads us to a new conservation law. Consider the case when there
are no external forces (Fj(ext) = 0). Eq. (6.10) becomes
I
N
pO (t2) =
j =1
mj Ivj/O (t2) =
N
j =1
mj Ivj/O (t1) = IpO (t1).
(6.11)
194
CHAPTER SIX
I
I
ey
ey
O
ex
Q
O
ex
I
vQ/O
I
I
vP/O
FQ,P
Q
g
P
–mQgey
vQ/O
P
FP,Q
–mPgey
P
(a)
Tey
Q
(b)
(c)
Figure 6.3 Sticky impacts.
In other words, in the absence of external forces, the total linear momentum of
all the particles stays the same throughout their trajectories, regardless of the forces
between each pair. This is a profound result and a broader statement of conservation
of linear momentum than in Chapter 4; it proves very useful in the next section and
throughout the book. It is so important that we elevate it to a new law.
Law 6.1 The law of conservation of total linear momentum states that, if the total
external force is zero, the total inertial linear momentum of a system of particles
is a constant of the motion.
Note that this law is a direct consequence of Newton’s third law and is true for
any internal forces acting among the particles. Thus total linear momentum can be
conserved even if there are forces acting on the individual particles as long as the total
external force is zero.
Example 6.3 Sticky Impact
Consider particles P and Q, as shown in Figure 6.3a. Suppose the particles collide and
stick together to form a new particle labeled PQ. This is a good model of a collision
between two cars, for instance. Since there are no external forces, we can use the
conservation of total linear momentum in Eq. (6.11) to solve for the final velocity of
PQ. From conservation of total linear momentum, we have
mP IvP/O (t2) + mQIvQ/O (t2) = mP IvP/O (t1) + mQIvQ/O (t1).
Using IvPQ/O (t2) = IvP/O (t2) = IvQ/O (t2),
I
vPQ/O (t2) =
mP IvP/O (t1) + mQIvQ/O (t1)
mP + m Q
,
which completes the example. We found the final velocity of the joint particle PQ in
terms of the initial velocities of particles P and Q regardless of what the force between
P and Q might be, as long as they collide and stick (we treat bouncy collisions in
Section 6.2).
LINEAR MOMENTUM OF A MULTIPARTICLE SYSTEM
195
Example 6.4 Sticky Impact with a Pendulum
Consider particles P and Q, as shown in Figure 6.3b, where P is now attached to
a pendulum and gravity is present. Suppose Q collides and sticks to P , which is
initially at rest. Let us determine the velocity of the joint particle PQ after the collision
using the velocity of Q before the collision. This is an example of a problem that
requires some care, since the assumptions used to find Eq. (6.11) no longer apply.
We cannot use the conservation of total linear momentum in Eq. (6.11) because the
pendulum rod exerts an external force on P and gravity exerts an external force on
both particles. Nonetheless, using the same approach that led to Eq. (6.11) will show
that the horizontal component of momentum is conserved during the collision.
Let t1 be the time of impact and t2 be the short time later when the particles are
firmly stuck together. The integral form of Newton’s second law, Eq. (4.2), for each
particle is
I
I
t2
mP vP/O (t2) = mP vP/O (t1) +
FP(ext)dt + FP ,Q(t1, t2)
t1
mQIvQ/O (t2) = mQIvQ/O (t1) +
t2
t1
(ext)
FQ
dt + FQ,P (t1, t2).
We also have that IvP/O (t1) = 0 and IvPQ/O (t2) = IvP/O (t2) = IvQ/O (t2). By
Newton’s third law, the linear impulses between the particles are equal and opposite.
Therefore, adding these two equations together yields
(mP + mQ)IvPQ/O (t2) = mQIvQ(t1) +
t2
t1
(ext)
(FP(ext) + FQ
)dt.
From the free-body diagram Figure 6.3c, the external forces on particles P and Q
(ext)
= −mQgey .2 Let ẋ and
during the collision are FP(ext) = (T − mP g)ey and FQ
ẏ denote the Cartesian coordinates for speed in the inertial frame I . That is, let
Iv
PQ/O = ẋPQ ex + ẏPQ ey . In terms of frame I components we have
(mP + mQ)ẋPQ(t2) = mQẋQ(t1)
t2
(mP + mQ)ẏPQ(t2) = mQẏQ(t1) +
(T − (mP + mQ)g)dt.
t1
We solve for ẋPQ(t2) from the first equation. In fact, this equation is the horizontal
component of the law of conservation of total linear momentum. (The second equation
could be used to find the magnitude of the tension force, which would be useful for
sizing the rod if we were building such a system.) We also have the constraint imposed
by the pendulum rod on the final velocity of particle PQ,
ẏPQ(t2) = 0.
2 The assumption that the collision is very short implies that the pendulum force T on particle P stays
approximately vertical during the collision. In reality, it will have a very small horizontal component as
well.
196
CHAPTER SIX
i
I
e2
ri/G
G
ri/O
rG/O
O
e1
e3
Figure 6.4 Center of mass G of a collection of many particles.
In words, this constraint states that, immediately after the collision, the joint particle
can only move horizontally. Now we know the velocity of particle PQ immediately
after the collision:
I
vPQ/O (t2) = ẋPQ(t2)ex =
mQ
mP + m Q
ẋQ(t1)ex ,
(6.12)
which depends only on the particle masses and the initial horizontal speed of particle Q.
6.1.3 The Center of Mass
This subsection introduces the concept of the center of mass of a collection of
particles. The center of mass leads to important conclusions about the motion of a
collection of particles. The center of mass also becomes extremely important later
when treating the angular momentum of a multiparticle system and when treating
rigid bodies in Part Three.
Definition 6.2 The position of the center of mass rG/O of a collection of N
particles, as shown in Figure 6.4, is the vector sum of the mass-weighted particle
positions:
1
rG/O =
mG
N
where mG =
!N
i=1 mi
mi ri/O ,
(6.13)
i=1
is the total mass of all particles.
A consequence of this definition is that the mass-weighted position of each particle
relative to the center of mass is zero. To see this, start with the vector triad
ri/O = rG/O + ri/G.
(6.14)
LINEAR MOMENTUM OF A MULTIPARTICLE SYSTEM
197
Using the definition of the center of mass, we find
N
N
mGrG/O =
mi (rG/O + ri/G) = mGrG/O +
i=1
mi ri/G,
i=1
which implies
N
mi ri/G = 0.
(6.15)
i=1
We will use this result a lot! We call it the center-of-mass corollary.
Motion of the Center of Mass
Why is the concept of the center of mass helpful? One reason is that, in the absence
of external forces, the center of mass is fixed (or moves at a constant velocity) in an
inertial frame. To see this, return again to Eq. (6.1). Instead of using the integrated
form, we can sum Newton’s second law for each particle over all of the particles to get
N
mj
j =1
Id 2 dt 2
N
Fj(ext),
rj/O =
j =1
where, as before, the double sum over the internal forces is zero. Again we substitute
for rj/O from the vector triad in Eq. (6.14):
mG
Id 2 dt 2
rG/O +
N
mj
j =1
Id 2 rj/G =
dt 2
N
Fj(ext),
j =1
!N
where we have used the fact that mG =
j =1 mj . Using the center-of-mass corollary
in Eq. (6.15), the second term evaluates to
N
Id 2 Id 2 N
mj rj/G = 0.
mj 2 rj/G = 2
dt j =1
dt
j =1
=0
This calculation yields the important result,
Id
dt
N
I
(ext)
,
Fj(ext) =
FG
pG/O =
(6.16)
j =1
where we have introduced a new quantity,
I
mG
pG/O =
Id dt
rG/O .
(6.17)
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CHAPTER SIX
Eq. (6.17) looks like the linear momentum of a single particle of mass mG located at
the center of mass G. Comparing to Eq. (6.8), we observe that IpG/O is equal to the
total linear momentum IpO of the collection of particles.
What have we found? Summing Newton’s second law over all particles yields a
single equation of motion for the center of mass of the collection. The vector sum of
all external forces acts as if it were a single force applied to a single particle located at
the center of mass. In other words, to study the translational dynamics of a collection
of particles, we can treat the collection as a single equivalent particle of mass mG
located at the center of mass G and ignore the internal forces! The center of mass
follows a trajectory determined by solving Newton’s second law independently of the
relative motion of the particles.
This result justifies many of the examples done so far in which we treated the translational motion of an extended body as if it were a point mass. It also gives us a new
way of thinking about a multiparticle system. For many problems, the motion of the
particles relative to the center of mass is of more interest than that of the center of mass
itself. (Consider, e.g., the motion of astronauts relative to the center of mass of the
space station as the station orbits the earth.) This result shows that we can separate the
motion of the center of mass from the motion of the system about the center of mass.
(ext)
= 0, then IpG/O is constant, which is another
Eq. (6.16) also implies that, if FG
statement of the law of conservation of total linear momentum. In the absence of
external forces—or if the external forces sum to zero—the center of mass of the
collection travels at a constant velocity. Keeping track of the center of mass can thus
lead to important insights or act as an important check to avoid mistakes.
Finally, we could easily write the impulse form of Eq. (6.16), as done in Chapter 4
for the momentum of a single particle. All results in Chapter 4 describing an impulse
on a single particle apply here to the center of mass of a particle collection treated as
a single particle.
Example 6.5 A Falling Collection of Particles
Consider the collection of particles falling in a uniform gravitational field (with
acceleration gey ), as shown in Figure 6.5a. Each particle has a force −mi gey acting on
it, as shown in the free-body diagram Figure 6.5b. Thus each particle obeys Newton’s
second law,
mi
ri/O = Fi = −mi gey .
Id 2 dt 2
What is the equation of motion of the center of mass? From Eq. (6.16) we consider
the sum
Id dt
pG/O =
I
N
−mi gey ,
i=1
which simplifies to
mG
Id 2 dt 2
rG/O = −mGgey .
LINEAR MOMENTUM OF A MULTIPARTICLE SYSTEM
199
i
I
ey
ri/G
i
ri/O
G
–migey
rG/O
G
O
ex
(a) Coordinates and reference frame
(b) Free-body diagram
Figure 6.5 A collection of particles falling freely in a uniform gravitational field.
The center-of-mass equation of motion is
Id 2
dt 2
(rG/O ) = −gey .
The center of mass constantly accelerates at g in the −ey direction. We treated
the collection of particles as a single one located at the center of mass and found the
equation of motion for it. This is possible whether the particles are free, as in this
example, or connected in a flexible or rigid body.
This example shows that the acceleration of the center of mass is independent of
the arrangement of particles. It is a function of the center-of-mass position only (here,
the acceleration is a constant). Unfortunately, that is not always the case. Sometimes
the force on the center of mass depends on the relative position of the particles,
coupling the motion of the center of mass to the motion relative to the center of mass.
Tutorial 6.1 discusses such a situation.
Example 6.6 The Center of Mass of an Overhead Crane
Consider again the crane of Example 6.2. Suppose the system is released from rest at
x(0) = 0 and θ (0) = 0. This example examines the motion of the center of mass G
of the crane.
Using the kinematics of the pendulum and the block derived in Example 6.2 and
Definition 6.2, we can compute the position of the center of mass:
rG/O =
mQxex + mP (xex + ler )
mG
mP
= xex +
ler .
mG
200
CHAPTER SIX
Let l˜ =
mP
mG l.
The kinematics of G are
˜r
rG/O = xex + le
vG/O = ẋex + l˜θ̇eθ
I
aG/O = ẍex + l˜θ̈eθ − l˜θ̇ 2er .
I
Note the similarities between the kinematics of G and P . In fact, in the limit mQ → 0,
the kinematics of G and P are identical.
Using Eq. (6.16) and the free-body diagram Figure 6.2c, we find that
Id
dt
pG/O = mG(ẍex + l˜θ̈eθ − l˜θ̇ 2er ) = (N − mGg)ey
I
(6.18)
Eq. (6.18) yields two scalar equations in terms of three unknowns: ẍ, θ̈, and N . Therefore, we cannot directly solve this algebraic system of equations without introducing
another equation, such as Eq. (6.15). Nevertheless, we can gain insight into the motion of the system from the observation that the total external force acts only in the
vertical direction.
In the absence of horizontal external forces, the horizontal component of IpG/O
is conserved. And, since the particles are initially at rest, the horizontal component
of IpG/O is zero for all time and the center of mass does not move horizontally. We
express this observation mathematically by writing IpG/O in terms of components in
the I frame, using Eq. (6.2):
pG/O = mG(ẋ + l˜θ̇ cos θ)ex + mGl˜θ̇ sin θey .
I
We have the conservation equation
ẋ + l˜θ̇ cos θ = 0,
which can be integrated by inspection to find
x = −l˜ sin θ + x(0).
(6.19)
Even without solving for the equations of motion of x and θ, we found that their
trajectories must satisfy Eq. (6.19). If |θ| < π/2, this equation shows that, as the
pendulum P swings one way, the block Q slides the other way. In fact, as you might
expect, P oscillates like a pendulum; Eq. (6.19) implies that Q must also oscillate in
such a way as to keep the horizontal position of G fixed.
Motion Relative to the Center of Mass
To complete the picture of a multiparticle system, we write the equations of motion
for the position of each particle relative to the center of mass, which was examined in
Section 3.6. We attach a nonrotating frame B to the center of mass, as in Figure 6.6,
and consider the trajectory ri/G of each particle in this frame. The equations of motion
for each particle relative to B are given by Eq. (3.57), with P replaced by i, O LINEAR MOMENTUM OF A MULTIPARTICLE SYSTEM
B
201
b2
i
I
e2
ri/G
ri/O
G
b1
rG/O
O
e1
e3
Figure 6.6 Center of mass G of a collection of many particles with a nonrotating frame B
attached to G.
replaced by G, and the force Fi separated into external and internal components:
mi Bai/G = Fi(ext) +
N
Fi,j − mi IaG/O .
j =1
We can also use the angular-momentum form of Newton’s second law discussed
in Section 4.2.2 to find equations of motion for each particle relative to the center of
mass. The result is given by Eq. (4.15), with P replaced by i and Q replaced by G:
Id
dt
hi/G = Mi/G − ri/G × mi IaG/O ,
I
(6.20)
where
Mi/G = ri/G × Fi(ext) + ri/G ×
N
Fi,j .
j =1
Chapter 7 examines Eq. (6.20) in more detail.
We now have equations of motion for the trajectory of the center of mass and for
the relative position of each particle about the center of mass. It may seem like we
have more equations than degrees of freedom, but the inclusion of Eq. (6.15) as a
constraint reconciles this imbalance. Chapter 7 returns to the concept of separating
the motion of the center mass from that of the particles relative to the center of mass.
It is also worth noting that, in the absence of external forces, when the linear
momentum of the center of mass is a constant (i.e., it is stationary or moves at
constant velocity), Newton’s second law for each particle relative to the center of
mass reduces to
B
ai/G =
N
1 Fi,j .
mi j =1
(6.21)
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CHAPTER SIX
m1
I
B
b2
ey
k
G
r1/O
m2
rG/O
b1
r2/O
O
ex
Figure 6.7 Two particles sliding in the plane connected by a linear spring with spring
constant k.
Example 6.7 Two Sliding Masses Connected by a Spring
Consider two particles of equal mass connected by a linear spring with spring constant
k and rest length l0. The masses are constrained to slide on a frictionless table, as
shown in Figure 6.7. This is a four-degree-of-freedom problem: each particle can
move in two directions on the table. The only external force is gravity (directed into the
page), which is countered by the normal force from the table (directed out of the page).
In this example we find the equations of motion for the center of mass of the system
and for the motion of the particles relative to the center of mass after one particle is
struck by an impulse F1(t1, t2), assuming the system begins at rest.
Because there are only two masses, the center of mass is always on the line joining
them. From Eq. (6.16) the equation of motion of the center of mass is
Id 2 dt 2
(ext)
rG/O = FG
.
Using Eqs. (4.2) and (6.16) with t ≥ t2, we find
I
vG/O (t) =
1
F1(t1, t2).
mG
(6.22)
Thus, our results from Chapter 4 show that the center of mass travels at constant
velocity in the direction of the impulse.
For motion relative to the center of mass after the impulse we use Eq. (6.21) and
the vector triads r1/O = rG/O + r1/G and r2/O = rG/O + r2/G to obtain
B
a1/G =
B
a2/G =
F1,2
m1
F2,1
m2
=
−k r1/G − r2/G − l0 r̂1/2
m1
=
−k r1/G − r2/G − l0 r̂2/1 ,
m2
where l0 is the rest length of the spring and, as usual, the spring force acts only along
the line connecting the particles and is proportional to the compression (or extension)
LINEAR MOMENTUM OF A MULTIPARTICLE SYSTEM
203
of the spring. The unit vector directed from particle 2 to particle 1 is given by
r1/O − r2/O
r̂1/2 =
r1/O − r2/O =
r1/G − r2/G
r1/G − r2/G
= −r̂2/1.
Because there are no external forces, the frame B attached to the center of mass is
also an inertial frame.
These equations, combined with the center-of-mass equation of motion, provide
six scalar equations of motion for the system. However, the system has only four degrees of freedom. Thus there must be two scalar constraint equations. The constraints
are given in vector form by the center-of-mass corollary (Eq. (6.15)):
m1r1/G + m2r2/G = 0.
For this simple problem it is perhaps more interesting to find an equation of motion
for the relative position between the particles, r2/1 = r2/G − r1/G, thus eliminating
the need to carry the center of mass constraint. This is done by subtracting the two
equations of motion relative to the center of mass:
Bd 2 dt 2
r2/1 = Ba2/G − Ba1/G
=
−k(m1 + m2)
(r2/1 − l0r̂2/1).
m1m2
(6.23)
The quantity m1m2/(m1 + m2) is often called the reduced mass.
Eq. (6.23) has the form of an equation for
simple harmonic motion; the particle
separation will oscillate at a frequency ω = k(m1 + m2)/m1m2, while the center of
mass travels at the constant velocity given by Eq. (6.22). For multiparticle systems
such as this one we often associate the degrees of freedom with what are called
the modes of motion, rather than the trajectories of the individual particles. In this
problem, the four modes are the two translational degrees of freedom of the center of
mass, the rotation of the line connecting the particles about the center of mass, and
the oscillation of the particles along the line connecting them.
The initial conditions for the oscillation are also found from the impulse equation.
Since particle 1 is hit by the impulse, we have
I
v1/O (t2) =
1
F1(t1, t2)
m1
I
v2/O (t2) = 0
or, in terms of the relative velocity,
B
v2/1(t2) = −
1
F1(t1, t2).
m1
This is another example where the motion relative to the center of mass is completely independent of the motion of the center of mass. In this case, it is because there
are no external forces, so the linear momentum of the center of mass is conserved.
204
CHAPTER SIX
Figure 6.8 Deflecting an asteroid using an explosive to split it apart.
Example 6.8 Deflecting an Asteroid
Consider the inverse of Example 6.3: rather than two objects colliding and sticking,
we have two objects traveling together at a constant velocity before bursting apart.
This is a simple model for myriad phenomena, from explosions, to propulsion, to
the recoil in a gun. One particularly interesting application is asteroid deflection,
shown in Figure 6.8. There is rising concern that some time in the next few centuries
a large asteroid may collide with the earth, causing devastating damage. To avoid
this scenario, some scientists have proposed setting off a large explosive (probably
nuclear) on the surface of the asteroid to redirect it away from the earth.
Consider again two particles P and Q that are initially joined and traveling at
mP + mQ .
the velocity IvPQ/O (t0). The total mass of the joint particle PQ is mPQ =
B
3
Suppose particle Q is ejected with relative velocity vQ/P , where B is a nonrotating
frame fixed to the center of mass G. What is the final velocity of particle P ?
Since no external forces are acting, total linear momentum is conserved and G
travels at a constant velocity IvPQ/O (t0). From conservation of total linear momentum,
we have
mQIvQ/O (tf ) + mP IvP/O (tf ) = mPQIvPQ/O (t0),
(6.24)
where tf is some time after P is ejected. Differentiating the vector triad
rQ/O = rQ/P + rP/O
yields the relative motion:
vQ/O (tf ) = IvQ/P (tf ) + IvP/O (tf )
I
= BvQ/P (tf ) + IvP/O (tf ).
The latter equality holds because frame B is not rotating (see Section 3.6). Substituting
Iv
Q/O (tf ) into Eq. (6.24), we find
mQBvQ/P (tf ) + mPQIvP/O (tf ) = mPQIvPQ/O (t0),
which can be solved for the final velocity of particle P ,
3 The
ejected particle Q represents the portion of the asteroid broken off by the explosive.
LINEAR MOMENTUM OF A MULTIPARTICLE SYSTEM
I
vP/O (tf ) = IvPQ/O (t0) −
mQ
mP + m Q
205
B
vQ/P (tf ).
Normally mQ mP , so the ejected particle makes only a slight change to the
velocity of the remaining mass. Fortunately, if caught early enough, only a very small
change in the asteroid velocity is needed to cause it to miss the earth.
There is one caveat that should be kept in mind here. Although we have rigorously
shown that we can combine the particles and write a single equation of motion for
the center of mass in terms of the net force on it, it is not always the case that we can
solve this equation without also considering the relative motion of the particles. This
(ext)
is because in many common situations the net force FG
cannot be determined as a
function of the center-of-mass position only but may (and often does) depend on the
positions (or velocities) of the individual particles. This situation produces a coupling
that can’t always be reconciled. Tutorial 6.1 illustrates such a situation.4
6.2 Impacts and Collisions
This section examines one of the most commonly encountered dynamics problems:
collisions. Be it cars crashing, balls bouncing, baseball bats hitting home runs, or
many other phenomena, we experience two objects colliding with each other on
a regular basis. Our goal is to understand the dynamics of the objects during and
after the collisions. Remarkably, despite its ubiquity, the collision is one of the
most difficult dynamic problems to model. Accurately representing the complete
physics would require detailed modeling of the shapes of the objects, their material
properties, the deformations each undergoes during contact, and any energy losses
during deformation. This task is clearly formidable and certainly beyond our scope.
However, all is not lost. By making a few simplifying assumptions, we can use
the tools acquired so far—namely, linear impulse and the conservation of total linear
momentum—to form remarkably effective predictive models of many impacts.
6.2.1 Planar Collision between Two Particles
The overall objective of this subsection can be summed up quite succinctly: given the
velocities of two particles before a collision, find the velocities of the two particles
after the collision.
This objective is shown schematically in Figure 6.9. Although easy to state, the
task is difficult to execute. In fact, it is hopelessly complex without first making a few
simplifying assumptions. Even though the following list of assumptions may seem
rather restrictive, they are, in fact, met by a wide variety of situations. For some cases
4 It is an interesting historical footnote that this problem prevented Newton from publishing his laws of
dynamics and the differential calculus for more than 30 years (thus ceding the notation to Liebnitz). He
found that he could predict the motion of the moon about the earth as long as he treated each as a point
mass. Proving it was okay to ignore the distribution of mass in each was a formidable problem and required
the invention of the integral calculus. As it turns out, it is only correct to treat the gravitational force as
coming from a point mass if the earth is a sphere. For any other shape, the relative orientation of the earth
and the moon affects the trajectory of the moon’s center of mass.
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CHAPTER SIX
I
e2
Q
I
I
vQ/O
P
I
vP/O
P
O
vQ/O
Q
I
vP/O
e1
Figure 6.9 A collision between two particles changes both the magnitude and direction of the
velocity of each particle.
when one or more of the following assumptions are not met, we can often apply the
general approach developed below, but the specific results are not guaranteed to hold.
Assumption 6.1 The collision occurs in a very short time interval, during which the
objects do not move significantly. We thus do not need to model the details of the
force acting between the two objects during the impact; instead, the details are
lumped into the linear impulse imparted to each object. An implication of this
assumption is that the collision instantaneously changes the particle velocities.
Assumption 6.2 There are no external forces acting on either particle during the
collision, which implies that the total linear momentum is conserved. If there
are no external forces acting on either particle at any time, then their trajectories
before and after the collision are straight lines.
Assumption 6.3 The colliding objects are infinitesimally small particles.We model
each object as an infinitesimally small sphere of uniform density located at
its center of mass. An implication of this assumption is that the pre-collision
trajectories are such that the collision would occur no matter how small the
objects are.
Assumption 6.4 There are no frictional forces between the colliding objects. An
implication of this assumption and the previous one is that the (internal) forces
generated by the collision act along the line through the center of mass of each
body.5
Chapter 12 examines a more general model of impacts in which Assumption 6.3
is relaxed. Note that sticky collisions in which particles collide and stick (as in
Example 6.3) violate Assumption 6.4.
For brevity, we continue our two-dimensional treatment when studying collisions
and only treat impacts that occur in a plane. However, the analysis is completely
general and extends to three dimensions. The remainder of this section develops a
5 This line is called the line of impact and our model of a collision is sometimes called a central impact
collision.
LINEAR MOMENTUM OF A MULTIPARTICLE SYSTEM
207
Q
Q
I
vQ/O
P
I
vP/O
I
Q
I
vP/O
P
(a)
I
vQ/O
(b)
vQ/O
I
Q
Q
P
I
vP/Q · en = 0
(c)
v Q/O
I
P
I
vP/O
(d)
vP/O
P
(e)
Figure 6.10 The five phases of the collision of two particles. (a) Initial phase before the
collision; (b) start of deformation phase; (c) compressed phase; (d) end of restitution phase;
(e) final phase after the collision.
model of the impact between two bodies following the process outlined in Chapter 3;
that is, we define the appropriate reference frames and coordinate systems, draw freebody diagrams, and work out the particles’ motion using Newton’s laws.
Modeling the Impact
First we describe the physical model of the collision between two spherical particles,
labeled P and Q, using Assumptions 6.1–6.4. If the particles collide and stick
together, then Assumption 6.4 is violated; we find the final velocities using the
law of conservation of total linear momentum in Eq. (6.11). If the particles collide
and bounce apart, then we need a more detailed model. In fact, the final equations
developed for the bouncy collision are not valid for the perfectly sticky collision
(because of Assumption 6.4), as we shall see.
Using Assumption 6.1 (that the collision occurs over a very short, but finite, time
interval) we label the start and end times of the impact, respectively, by t1 and t2.
The particle velocities just before the collision are IvP/O (t1) and IvQ/O (t1) and the
final velocities are IvP/O (t2) and IvQ/O (t2). For the short time during the collision,
the particles squish together and then spring apart.6 Let tc denote the time at which
the particles stop compressing and start expanding, where t1 < tc < t2. (At t = tc , the
normal components of the particle velocities are the same—we explain what we mean
by “normal” below.) The collision is thus divided into five sequential phases, as shown
in Figure 6.10: (a) the initial phase, t < t1, before the collision; (b) the deformation
phase, from t1 to tc , when each particle compresses; (c) the compressed phase, which
is the time tc of the maximum compression when there is no relative motion between
the particles along the line of impact; (d) the restitution phase, from tc to t2, when each
particle returns to its original shape; and (e) the final phase, t > t2, after the collision.
Next we define two frames of reference and draw the free-body diagram for each
particle so that we can apply Newton’s second law to the collision. Figure 6.11a shows
the inertial frame I = (O, e1, e2, e3) and the collision frame C = (O , en, et , e3). The
6 We recognize that there appears to be an inconsistency here. If all particles are point masses (Assumption 6.3), with no extent, as we insisted earlier in the application of Newton’s second law, then how can
they “squish”? In a detailed treatment, this would be a problem and each object would have to be treated
as a collection of particles (i.e., a nonrigid body). However, as we are ignoring the details of the motion
during the impact, we can safely ignore this problem. Before and after the impact we still treat the bodies
as non-squishy particles.
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CHAPTER SIX
C
I
et
en
Q
e2
O′
P
FQ,P = FQ,Pen
Q
O
e1
FP,Q = –FQ,Pen
(a) Reference frames
P
(b) Free-body diagrams
Figure 6.11 Reference frames and free-body diagrams for the collision of two particles.
collision frame is defined by the point of impact O and the two unit vectors, et and
en, which lie in the same plane as the particle velocities. Because we have assumed
the particles are small spherical particles (Assumption 6.3), O always lies on the line
of impact connecting the centers. For large, and unequal sized, particles, finding O and the normal unit vector en involves consideration of the trajectory of both particles
and their relative positions. Chapter 12 treats this more complex case.
The situation simplifies greatly for small particles, however. If the particles have
no extent and collide, then the line of impact is parallel to the velocity of P relative to
the velocity of Q; that is, en is parallel to IvP/Q(t1) = IvP/O (t1) − IvQ/O (t1).7 Since
en has unit length, we have
en =
Iv
P/Q (t1)
I
vP/Q(t1)
= Iv̂P/Q(t1),
(6.25)
where the caret denotes a vector of unit length. The tangent unit vector completes the
frame: et = e3 × en.
Figure 6.11b shows the free-body diagrams for the two particles during the collision. According to Assumptions 6.2–6.4, the only force on each particle during the
collision is from the other particle; this force acts along the line of impact and is not
due to friction. The internal force between P and Q pushes the particles apart during
the collision. The direction of the collision force is fixed along the line of impact,
but its magnitude varies during the deformation and restitution phases. We write the
force on particle P due to particle Q as FP ,Q. By Newton’s third law, FP ,Q = −FQ,P .
According to the definition of frame C shown in Figure 6.11a, FP ,Q acts in the −en
direction.
Figure 6.12 shows a schematic plot of the strength of the force during the collision.
FP ,Q ≥ 0 denote the magnitude of FP ,Q and likewise for FQ,P . By
Let FP ,Q =
Newton’s third law, FP ,Q = FQ,P . Both forces satisfy the criteria for a linear impulse,
7 Why? Since the colliding particles are infinitesimally small spheres, then the line of impact is parallel to
rQ/P (t1), and rQ/P (t1) must be parallel to IvP/Q (t1) or else the particles would not collide. (See Chapter 12.)
FP,Q
LINEAR MOMENTUM OF A MULTIPARTICLE SYSTEM
–
FP,Q(t1, tc)
0
t1
209
–
FP,Q(tc, t2)
tc
t2
t
Figure 6.12 Magnitude FP ,Q of the interaction force versus time. The area under the curve
between t1 and tc is the magnitude of the deformation linear impulse. The area under the curve
between tc and t2 is the magnitude of the restitution linear impulse. The deformation and
restitution linear impulses are not necessarily the same. (The restitution impulse cannot be the
larger of the two because of energy considerations.)
since we assume that t =
t2 − t1 is very short. The linear impulse imparted to particle
P from the collision with particle Q is FP ,Q(t1, t2), which has magnitude F P ,Q(t1, t2)
and direction −en. We distinguish between the linear impulse FP ,Q(t1, tc ) during the
deformation phase and the linear impulse FP ,Q(tc , t2) during the restitution phase.
By definition, the magnitude of the deformation linear impulse on particle P is the
area under the force magnitude curve during the deformation phase. (The analogous
statement holds for the restitution phase.) Just like the interaction forces, the linear
impulses on P and Q are equal in magnitude and opposite in direction.
Conservation of Linear Momentum
To solve for the velocities after the collision, we begin with the impulse form of
Newton’s second law in Eq. (4.2), which yields
mP IvP/O (t2) = mP IvP/O (t1) + FP ,Q(t1, t2)
(6.26)
mQIvQ/O (t2) = mQIvQ/O (t1) + FQ,P (t1, t2).
(6.27)
Letting u and v denote the components of the particle speed in the collision frame
C, we can write the velocities as
I
vP/O (t) = uP (t)et + vP (t)en
(6.28)
I
vQ/O (t) = uQ(t)et + vQ(t)en.
(6.29)
Thus, written as components in frame C, Eqs. (6.26) and (6.27) become
mP uP (t2)et + mP vP (t2)en = mP uP (t1)et + mP vP (t1)en − F P ,Q(t1, t2)en
mQuQ(t2)et + mQvQ(t2)en = mQuQ(t1)et + mQvQ(t1)en + F Q,P (t1, t2)en.
(Recall that FP ,Q = FQ,P and act in the opposite directions; hence the choice of signs
above.)
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Since there is no linear impulse in the tangential direction, the et component of
linear momentum is conserved for each particle:
uP (t2) = uP (t1)
(6.30)
uQ(t2) = uQ(t1).
(6.31)
This observation, of course, implies that the tangential component of the total linear
momentum is also conserved during the collision, which we can see by adding
Eqs. (6.30) and (6.31).
For the normal direction, we have
mP vP (t2) = mP vP (t1) − F P ,Q(t1, t2)
(6.32)
mQvQ(t2) = mQvQ(t1) + F Q,P (t1, t2).
(6.33)
Since the linear impulses acting on P and Q are equal and opposite, the normal
component of total linear momentum is conserved. Adding Eqs. (6.32) and (6.33)
and using F P ,Q(t1, t2) = F Q,P (t1, t2), we obtain the conservation equation
mP vP (t2) + mQvQ(t2) = mP vP (t1) + mQvQ(t1).
(6.34)
Eq. (6.34) and the corresponding tangential-direction equation (obtained by adding
Eqs. (6.30) and (6.31)) are simply a rederivation of the conservation law for the total
linear momentum (see Eq. (6.11)) since, by Assumption 6.2, the only forces acting
during the collision are internal.
While Eqs. (6.30) and (6.31) allow us to trivially solve for the tangential components of the final velocities, Eq. (6.34) is a single equation with two unknowns,
vP (t2) and vQ(t2). To find another equation in these unknowns, we need to define a
new quantity, called the coefficient of restitution.
Coefficient of Restitution
The second equation needed to find vP (t2) and vQ(t2) from Eq. (6.34) comes from
using the integrated form of Newton’s second law in Eq. (4.2) for each particle again,
but this time we distinguish between the deformation and restitution phases:
mP IvP/O (tc ) = mP IvP/O (t1) + FP ,Q(t1, tc )
(6.35)
mQIvQ/O (tc ) = mQIvQ/O (t1) + FQ,P (t1, tc )
(6.36)
mP IvP/O (t2) = mP IvP/O (tc ) + FP ,Q(tc , t2)
(6.37)
mQIvQ/O (t2) = mQIvQ/O (tc ) + FQ,P (tc , t2).
(6.38)
By assumption, the two particles have the same normal speed at time tc : IvP/Q(tc ) .
en = 0. In addition, the impulses on P and Q are equal and opposite. (Note that, if
we add all four of these equations, we recover the conservation equation for total
linear momentum.) Nonetheless, we still do not have enough equations to solve for
LINEAR MOMENTUM OF A MULTIPARTICLE SYSTEM
211
vP (t2) and vQ(t2) because there is no explicit form for the deformation and restitution
impulses. To avoid having to do the detailed physics, we define a new quantity, the
coefficient of restitution.
Definition 6.3 The coefficient of restitution e of the collision between particles P
and Q is the ratio of the magnitudes of the restitution and deformation linear
impulses:
e=
F P ,Q(tc , t2)
F P ,Q(t1, tc )
=
F Q,P (tc , t2)
F Q,P (t1, tc )
.
The coefficient of restitution satisfies 0 ≤ e ≤ 1.
The coefficient of restitution is a joint property of the materials of the colliding
objects. It must be a non-negative number less than or equal to one to satisfy energy
conservation (the restitution force cannot do more work than the deformation force).
We use it to predict the behavior of objects after a collision. There are simple experiments for measuring e for different collisions without having to directly measure the
deformation and restitution forces (see, e.g., Example 6.12). Note that the coefficient
of restitution depends on the properties of the materials of both particles involved and
thus differs for every collision model.
It may seem like we haven’t gained much by defining e; however, we have actually
gained a great deal. This quantity gives a measure of the elasticity of the collision.
If e is one, then the collision is perfectly elastic; that is, all of the linear momentum
that deformed the bodies is “returned” during restitution. Conversely, if e is zero,
then the collision is perfectly inelastic and the two bodies remain together after the
collision (also called a plastic deformation, which violates Assumption 6.4). For
values between zero and one, the particles rebound with reduced normal velocities
(recall that we showed the tangential speeds were conserved). You may rightfully be
wondering where the energy goes when e < 1. If the particles leave with lower normal
speeds than before the collision, then energy must have been lost during the collision.
Indeed, the coefficient of restitution is, in one sense, a measure of the energy loss; the
excess energy goes into heating up or deforming the material. The deformation and
restitution forces are not conservative.
Using Eqs. (6.28), (6.29), (6.35)–(6.38), and Definition 6.3, we can solve for the
coefficient of restitution in terms of the various speeds:
e=
vP (t2) − vP (tc ) vQ(t2) − vQ(tc )
=
.
vP (tc ) − vP (t1)
vQ(tc ) − vQ(t1)
(6.39)
Note that the coefficient of restitution is independent of the masses and tangential
speeds of the two particles; it depends only on their speeds in the normal direction.
Eq. (6.39) is really a pair of equations. These two equations can be used to eliminate
the unknown intermediate speed, vP (tc ) = vQ(tc ). After a little algebra, we reach our
final expression:
e=
vP (t2) − vQ(t2)
vQ(t1) − vP (t1)
.
(6.40)
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CHAPTER SIX
In words, the coefficient of restitution is the ratio of the difference in magnitude of
the normal speeds before and after the collision.
Final Velocities
We now have two equations, Eq. (6.34) and Eq. (6.40), for the two unknowns, vP (t2)
and vQ(t2). Given the initial speed in the normal direction of each of the two particles
(i.e., vP (t1) and vQ(t1)), we can solve for the final speeds in terms of the coefficient
of restitution. Although it is usually easiest to use these two equations directly when
solving a problem, it is possible to find a general expression for the final velocities
by simultaneously solving the following pair of equations:
mP vP (t2) + mQvQ(t2) = mP vP (t1) + mQvQ(t1)
vP (t2) − vQ(t2) = e(vQ(t1) − vP (t1)).
(6.41)
(6.42)
By multiplying Eq. (6.42) by mQ and adding Eqs. (6.41) and (6.42) and, likewise,
multiplying Eq. (6.42) by mP and subtracting Eqs. (6.41) and (6.42), we find the final
normal speeds:
vP (t2) =
vQ(t2) =
mP − emQ
mP + m Q
mQ − emP
mP + m Q
vP (t1) +
vQ(t1) +
mQ(1 + e)
vQ(t1)
(6.43)
mP (1 + e)
vP (t1).
mP + m Q
(6.44)
mP + m Q
Combining Eqs. (6.43) and (6.44) for the normal direction with Eqs. (6.30) and (6.31)
for the tangential direction, we obtain the final velocities:
I
vP/O (t2) = uP (t1)et +
I
vQ/O (t2) = uQ(t1)et +
mP − emQ
mP + m Q
mQ − emP
mP + m Q
vP (t1) +
vQ(t1) +
mQ(1 + e)
vQ(t1) en
(6.45)
mP (1 + e)
vP (t1) en.
mP + m Q
(6.46)
mP + m Q
As noted earlier, this model does not hold for a perfectly plastic (i.e., sticky)
collision, for which e = 0. In that case, Eqs. (6.45) and (6.46) show that the normal
speeds of P and Q are equal but their tangential speeds are not! The no-friction
assumption in the tangential direction is violated for a perfectly sticky collision. For
such collisions, we use Newton’s second law and the law of conservation of total
linear momentum, as in Example 6.3. Eqs. (6.45) and (6.46) are only valid when the
coefficient of restitution e is greater than zero (that is, a bouncy collision).
Example 6.9 A Simple Bouncy Collision
A simple impact is where two objects of equal mass whose velocities are aligned
collide and rebound in a straight line, as shown in Figure 6.13. This is an excellent
LINEAR MOMENTUM OF A MULTIPARTICLE SYSTEM
P
I
I
Q
vP/O(t0)
213
I
P
vQ/O(t0)
Q
vP/O(tf)
(a)
I
vQ/O(tf)
(b)
Figure 6.13 Two equal-mass particles traveling in a straight line collide and bounce with
different velocities. (a) Before collision (t0 < t1). (b) After collision (tf > t2 ).
model of two pool balls colliding head-on. The tangential speeds are zero (uQ =
uP = 0). Using Eqs. (6.43) and (6.44) and the initial speeds vQ(t1) and vP (t1), the
final speeds of the two balls are
1
1
vP (t2) = (1 − e)vP (t1) + (1 + e)vQ(t1)
2
2
1
1
vQ(t2) = (1 − e)vQ(t1) + (1 + e)vP (t1).
2
2
Supposing the initial speed of Q is zero—corresponding, for instance, to the cue ball,
P , hitting a stationary ball, Q—the final speeds simplify to
1
vP (t2) = (1 − e)vP (t1)
2
1
vQ(t2) = (1 + e)vP (t1).
2
In this case, if e is one (a perfectly elastic collision), the linear momentum is completely transferred from P to Q: the cue ball stops, and the target ball moves off at
the initial speed of the cue ball. For e < 1, the coefficient of restitution is a measure
of the amount of speed that is transferred from one ball to the other.
Example 6.10 An Oblique Bouncy Collision
This example demonstrates how to use Eqs. (6.45) and (6.46) to find the final velocities
of two arbitrarily small particles P and Q after the oblique collision shown in
Figure 6.14a. We treat the collision as bouncy, that is, the coefficient of restitution e is
greater than zero. Since there are no external forces, the particle velocities are constant
before and after the collision. The main challenge is to write the initial velocities in
the collision frame.
We begin by assuming that the initial velocities are known:
I
vP/O (0) = ẋP (0)ex + ẏP (0)ey
(6.47)
I
vQ/O (0) = ẋQ(0)ex + ẏQ(0)ey .
We next use the particle velocities at impact to determine the tangential and normal
directions of the collision frame C = (O , en, et , ez ):
en =
Iv
P/Q (t1)
IvP/Q(t1)
=
Iv
P/O (0) −
Iv
Q/O (0)
IvP/O (0) − IvQ/O (0)
.
(6.48)
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CHAPTER SIX
Q
I
vQ/O
C
I
ey
et
I
Q
I
P
O
vP/O(t1)
en
vP/O
O′
I
ex
vQ/O(t1)
P
(a)
(b)
Figure 6.14 Oblique bouncy collision.
Since the normal unit vector makes a right-hand triad with et and e3 (the vector
pointing out of the collision plane), we have
et = ez × en.
(6.49)
Using Eqs. (6.48) and (6.49), we solve for the tangential and normal unit vectors from
the initial particle velocities in terms of the unit vectors ex and ey :
ṡ(0)en = ẋP/Q(0)ex + ẏP/Q(0)ey
ṡ(0)et = −ẏP/Q(0)ex + ẋP/Q(0)ey ,
(6.50)
2
2 . To write the initial
+ ẏP/Q
where xP/Q = xP − xQ, yP/Q = yP − yQ, and ṡ = ẋP/Q
particle velocities in the collision frame, we invert Eq. (6.50) to find
1 ẋP/Q(0)en − ẏP/Q(0)et
ṡ(0)
1 ey =
ẏP/Q(0)en + ẋP/Q(0)et .
ṡ(0)
ex =
(6.51)
Eq. (6.51) allows us to write the initial velocities (Eq. (6.47)) in the collision frame
C. We can then use Eqs. (6.45) and (6.46) to solve for the final velocities in frame C.
Finally, using Eq. (6.50), we can write the final velocities as components in frame I .
6.2.2 Collision between a Particle and a Surface
This section examines the (bouncy) collision between a particle and a surface. The
goal is similar to that of the previous section: given the velocity of the particle before
a collision with the surface, find the velocity of the particle after the collision.
The presence of an external force holding up the wall means that the total linear
momentum of the particle-surface system is not conserved. As a result, we cannot
directly use the results of Section 6.2.1 to study a surface impact. Instead, we must go
back to first principles. In doing so, we follow steps similar to those in Section 6.2.1
using the following simplifying assumptions:
LINEAR MOMENTUM OF A MULTIPARTICLE SYSTEM
215
P
P
I
P
P
P
vP/O
I
(a)
vP/O
(b)
I
I
vP/O(tc) · en = 0
(c)
vP/O
I
vP/O
(d)
(e)
Figure 6.15 The five phases of the collision of a particle and a surface. (a) Initial phase before
the collision; (b) start of deformation phase; (c) compressed phase; (d) end of restitution phase;
(e) final phase after the collision.
Assumption 6.1 The collision occurs over a very short time interval, during which
the particle does not move. Thus the collision instantaneously changes the
particle velocity.
Assumption 6.2 There are no external forces acting on the particle during the
collision, other than from the surface. For some cases where this assumption is
not met, we can still apply the same approach developed here, but the general
result is not guaranteed to hold (see Tutorial 6.2).
Assumption 6.3 The moving object is a spherical point mass; if it is an extended
body, we treat it as if it is a point mass located at the center of mass.
Assumption 6.4 There are no frictional forces between the particle and the surface. Thus the collision impulse acts along the line from the point of impact
through the center of the body.
Assumption 6.5 In the vicinity of the impact point, the surface is smooth, rigid,
and fixed; that is, the normal direction is well defined, the surface does not
compress, and there is some external force holding the surface in place.
To model the impact of particle P with a surface, we use the same terminology and
notation as in Section 6.2.1. We assume that there are five sequential phases during
the collision (see Figure 6.15): (a) the initial phase, t < t1, before the collision; (b) the
deformation phase, from t1 to tc , when P compresses; (c) the compressed phase, which
is the time tc of maximum compression of the particle; (d) the restitution phase, from
tc to t2, when P returns to its original shape; and (e) the final phase, t > t2, after P
t2 − t1 is very small.
leaves the surface. According to Assumption 6.1, t =
Next we describe the inertial and collision frames illustrated in Figure 6.16a. We
assume that the inertial frame is determined by the orientation of the surface near the
impact point and the (planar) trajectory of the particle. That is, let I = (O, ex , ey , ez ),
where the surface lies in the plane spanned by the unit vectors ey and ez , and the
trajectory of the particle lies in the plane spanned by the unit vectors ex and ey . Because
the position and orientation of the surface are fixed, there is a natural choice for the
collision frame. Namely, let C = (O , en, et , e3), where O is the point of impact,
et = ey , en = −ex , and e3 = −ez .
By Assumption 6.4, the only force acting on the particle during the collision is
the interaction force FP , which acts normal to the surface (in the en direction). We
denote the magnitude of the force by FP , so that FP = FP en. As in Section 6.2.1, we
split the collision impulse FP (t1, t2) = F P (t1, t2)en from the surface to the particle
216
CHAPTER SIX
et
P
P
C
O′
en
FP = FP en
ey
I
O
ex
(a) Reference frames
(b) Free-body diagram
Figure 6.16 Collision of particle P and a fixed surface.
into two components. The first, F P (t1, tc ), corresponds to the impulse during the
deformation phase and the second, F P (tc , t2), corresponds to the impulse during the
restitution phase. Figure 6.16b illustrates the interaction force FP . (The magnitude
of the interaction force and the deformation and compression impulses are still
adequately described by Figure 6.12 with F P ,Q replaced by F P .)
To solve for the final velocity of particle P after the collision, we begin again with
the impulse form of Newton’s second law:
mP IvP/O (t2) = mP IvP/O (t1) + FP (t1, t2)
= mP IvP/O (t1) + F P (t1, t2)en.
(6.52)
Let uP and vP denote the Cartesian coordinates for the speed of P in collision frame
C, that is, IvP/O = uP et + vP en. Eq. (6.52) becomes
mP uP (t2) = mP uP (t1)
(6.53)
mP vP (t2) = mP vP (t1) + F P (t1, t2).
(6.54)
We observe that the tangential component of the particle’s linear momentum is
conserved during the collision, but the normal component is not. Thus the final
tangential speed uP (t2) = uP (t1) is known, but the final normal speed vP (t2) is not.
To find vP (t2), we once again use the impulse form of Newton’s second law,
this time considering separately the deformation and restitution phases. Since the
deformation and restitution impulses act only in the normal direction, we only write
the normal component of Newton’s second law:
mP vP (tc ) = mP vP (t1) + F P (t1, tc )
(6.55)
mP vP (t2) = mP vP (tc ) + F P (tc , t2).
(6.56)
Eqs. (6.55) and (6.56) represent two equations in terms of four unknowns: vP (t2),
vP (tc ), F P (t1, tc ), and F P (tc , t2). Hence we need two more equations to algebraically
solve the system of equations. The first equation comes from the following intuitive
observation: the normal speed of the particle reverses direction at time tc . That is,
time tc denotes the instant in time before which the particle is still moving toward the
LINEAR MOMENTUM OF A MULTIPARTICLE SYSTEM
217
surface and after which the particle moves away from the surface. This means
vP (tc ) = 0,
(6.57)
which is our third equation.8 The final equation emerges by once again introducing a
coefficient of restitution.
Definition 6.4 The coefficient of restitution e of the collision between particle P
and a rigid, fixed surface is the ratio of the magnitude of the restitution linear
impulse to the magnitude of the deformation linear impulse:
e=
F P (tc , t2)
F P (t1, tc )
.
The coefficient of restitution satisfies 0 ≤ e ≤ 1.
Definition 6.4 essentially replaces the two unknowns F P (t1, tc ) and F P (tc , t2)
with a single unknown e. Recall that we assume the collision is bouncy, which means
that e > 0 (e = 0 corresponds to a sticky or plastic collision and e = 1 corresponds
to a perfectly bouncy or perfectly elastic collision). Using Eqs. (6.55)–(6.57) and
Definition 6.4, we can solve for the final normal speed:
vP (t2) = −evP (t1).
(6.58)
Using Eqs. (6.53) and (6.58), we find the final velocity in terms of the initial velocity.
The final velocity in terms of components in the collision frame is
I
vP/O (t2) = uP (t1)et − evP (t1)en.
(6.59)
Perhaps as important as Eq. (6.59) is Eq. (6.58), which yields a practical definition
of the coefficient of restitution in terms of the initial and final normal speeds:
e=−
vP (t2)
.
vP (t1)
(6.60)
Note that Eq. (6.60) is equivalent to Eq. (6.40) with vQ(t2) = vQ(t1) = 0. Thus the
coefficient of restitution of a collision between a particle and a fixed surface is the same
as that of a collision between a moving particle and a fixed particle. Example 6.12
describes a procedure for measuring the coefficient of restitution for a surface impact.
Example 6.11 A Carom in Billiards
Rather than two pool balls colliding, consider a carom, or bank shot off the rail, as
shown in Figure 6.17. The goal is to find the initial conditions—as a function of the
coefficient of restitution—that generate a rebound of the desired angle between the
ball and the rail. (We postpone considering rotation of the ball until Example 9.11.)
8 Note
the resemblance of this formula to its counterpart in the collision between two particles:
vP/Q (tc ) = 0.
218
CHAPTER SIX
I
vP/O(t2)
θ
I
vP/O(t1)
ϕ
P
Figure 6.17 A billiard ball impacting the rail at angle φ and banking off the rail at angle θ .
The incoming velocity is IvP/O (t1) = uP (t1)et + vP (t1)en. The incoming angle
with the rail is thus
tan φ =
−vP (t1)
.
uP (t1)
After the bank off the rail, the final velocity is given by Eq. (6.59). The bank
angle satisfies
tan θ =
−evP (t1)
= e tan φ.
uP (t1)
For a perfectly elastic collision, the ball leaves the rail at exactly the same angle as
it arrived (making bank shots much easier to align). However, the felt-covered rail
results in e < 1, reducing the rebound angle and making the carom more challenging.
Example 6.12 Measuring e for a Surface Impact
This example describes an experimental procedure for determining the coefficient of
restitution of the collision between a ball (particle P ) and the floor (the fixed surface).
The procedure is quite simple: we drop the ball from a known height and measure how
high it bounces. However, the presence of gravity violates Assumption 6.2, which
was used to derive Eq. (6.60). Thus we have to revisit the derivation of Eq. (6.60)
to determine the effect of gravity. We show that, even in the presence of gravity,
t2 − t1
Eq. (6.60) is a good approximation of e, as long as the bounce duration t =
is very short, where t1 and t2 denote the start and end times of the impact, respectively.
Let I = (O, ex , ey , ez ) and C = (O , en, et , e3), where I = C, as shown in Figure 6.18b. Consider the free-body diagrams of particle P shown in Figure 6.18c.
The only force acting on P before and after the collision is the force of gravity
−mP gey = −mP gen. The total force acting on P during the collision is the sum
of the force from the wall FP = FP en and the force of gravity. Since we drop the ball
from rest at time t = 0, the initial horizontal speed is zero and, since there is never
any force in the horizontal direction, the final horizontal speed is also zero.
We find the vertical speed of the particle at time t1 using conservation of energy,
since, prior to the impact, the only force acting on the particle is gravity (which is
conservative). When the particle is at height y0, it has potential energy mP gy0. Just
LINEAR MOMENTUM OF A MULTIPARTICLE SYSTEM
219
I=C
ey = en
y
P
y0
y3
–mPgey
g
–mPgey
FPen
y
O = O′
O
–mPgey
ex = et
(a) Coordinates
(b) Reference frames
(c) Free-body diagrams
Figure 6.18 Measuring e for a surface impact. Particle P is colored black before impact, gray
during impact, and white after impact.
before it hits the floor the height is zero, so all of the potential energy is converted to
kinetic energy. Conservation of energy yields
1
mP (vP (t1))2 = mP gy0,
2
which results in a speed before impact of
vP (t1) = − 2y0g.
(6.61)
We have used the negative root because the velocity is in the −en direction.
Let t3 > t2 denote the time at which the ball reaches its maximum bounce height y3.
We find the vertical speed of the particle at t2 the same way as we found the vertical
speed at t1, that is, by using energy conservation. Thus, if the particle reaches height y3
with potential energy mP gy3, then its initial speed after impact is
(6.62)
vP (t2) = 2y3g.
If we were to ignore the effect of gravity during the impact, we could find e by
substituting Eqs. (6.61) and (6.62) into Eq. (6.60). However, let us see how big an
effect gravity has on the previous analysis. As before, we use the impulse form of
Newton’s second law for both the deformation and restitution phases, which has the
normal components
mP vP (tc ) = mP vP (t1) + F P (t1, tc ) −
tc
mP gdt
(6.63)
mP gdt.
(6.64)
t1
mP vP (t2) = mP vP (tc ) + F P (tc , t2) −
t2
tc
Since mP and g are constant during the interval t1 to t2, the integrals in Eqs. (6.63)
and (6.64) evaluate trivially to the small quantities mP g(tc − t1) and mP g(t2 − tc ),
respectively. Let O(t) denote a small quantity that is less than or equal to a constant
220
CHAPTER SIX
times t. Since t1 < tc < t2, the integrals in Eqs. (6.63) and (6.64) are both O(t).
Using this observation, Eq. (6.57), and Definition 6.4, we find that
e=
vP (t2) + O(t)
vP (t2)
=
+ O(t).
−vP (t1) + O(t) −vP (t1) + O(t)
(6.65)
We now use the binomial expansion described in Example A.1 of Appendix A. That
is, the Taylor series expansion of the first term on the right-hand side in Eq. (6.65)
can be expanded using the fact that O(t) is a small quantity:
1
−1
≈
(1 + O(t)).
−vP (t1) + O(t) vP (t1)
This estimate lets us rewrite Eq. (6.65) as
e=−
vP (t2)
+ O(t).
vP (t1)
Thus ignoring gravity and using Eq. (6.60) produces an error proportional to the
impact duration t.
Using Eqs. (6.61) and (6.62) to substitute for the speeds, we find the coefficient of
restitution is
e=
y3
+ O(t).
y0
In the limit t → 0, e ≈ y3/y0, which is what we would have obtained using
Eqs. (6.61) and (6.62) in Eq. (6.60) directly. This approximation is only as good as
the assumption of a short impact duration.
6.3 Mass Flow
This section shifts from considering only two particles to the other extreme and
examines the behavior of a system with infinitely many particles. Parts Three and
Four examine rigid collections of many particles, or rigid bodies, in great detail. Here
we study the specialized problem of a variable-mass system, that is, a collection of
particles that is gaining or losing mass (or both) by means of particles flowing in
and/or out of the system. It is far beyond our scope to study fluid dynamics in any
detail. However, as with collisions, we need only make a few simplifying assumptions
to be able to solve a number of useful problems using only a particle-dynamics model.
All analyses of variable-mass systems begin with the concept of the control volume.
The control volume is simply a geometric shape that encloses the particles for which
we desire to find trajectories (see Figure 6.19). Recall how we approached the study
of multiparticle systems in Section 6.1. As long as we keep track of every particle,
we can use conservation of total linear momentum or the equations for linear impulse
and momentum on the total linear momentum of the entire collection. Here we isolate
a subset of the particles, draw a control volume around them, and examine the total
linear momentum of the particles in the control volume. Our goal is to find an equation
LINEAR MOMENTUM OF A MULTIPARTICLE SYSTEM
221
Control volume
Figure 6.19 An example of a control volume isolating a vessel and a collection of particles as
part of a stream.
of motion for a point fixed to the control volume. The only requirement is that we keep
track of the total linear momentum of the same collection of particles before and after
a particle is ejected or added.
In the two subsections that follow we study two situations: (a) a steady stream of
particles entering and leaving a system such that the total mass stays the same and
(b) a system that is ejecting or collecting particles such that its mass is increasing or
decreasing while it is moving.
6.3.1 Steady Streams
Consider a system consisting of a stream of fluid passing through an open vessel, as
shown in Figure 6.20. We take as our control volume the boundary of the vessel; it
thus contains the particles making up the vessel and the particles making up the fluid
passing through it. The control volume has center of mass G and attached body frame
(ext) . Our goal is to find an
B. The control volume is acted on by an external force Fcv
equation of motion for G. The system of particles we examine consists of the mass in
the control volume plus the small differential mass m in the stream about to enter
the control volume.
I
e2
B
(ext)
Fcv
O
Control volume
b2
(ext)
Fcv
Δm
b2
Iv
G/O
e1
G
Δm
B
b1
Bv
Q/G
Q
G
b1
P
Bv
P/G
(a)
(b)
Figure 6.20 A system of mass mG consisting of a steady stream of particles entering and exiting a
control volume. (a) t = t1. (b) t = t2 = t1 + t.
222
CHAPTER SIX
A general solution to this problem involves sophisticated fluid mechanics. We
therefore make a few restrictive assumptions, as we did for collisions, that allow us
to use a particle model to find the equations of motion. Although many problems can
be solved with these assumptions, it is important to remember them to avoid using
the resulting equation on problems for which it doesn’t apply.
Assumption 6.6 The particles making up the fluid are evenly distributed in the
control volume and maintain the same distribution throughout the motion of G.
Thus the center of mass G has a fixed position in the control volume.
Assumption 6.7 The fluid maintains a constant velocity relative to the frame B
fixed to the control volume throughout the motion of G.
Assumption 6.8 Fluid mass enters the control volume at the same rate as it leaves,
so that the total mass of particles in the control volume is constant.
Assumption 6.9 The control volume does not rotate relative to the inertial frame
I . That is, we only study translational motion of the control volume.
Let IvG/O represent the inertial velocity of the center of mass of the control
volume (and thus also of the vessel or any other stationary mass in the control volume). The velocity of each fluid particle in the control volume relative to frame B is
given by Bvj/G. The differential mass about to enter the control volume has a velocity
relative to the control volume of BvP/G. The total linear momentum of the system of
particles consisting of the control volume and the mass P about to enter at time t1 is
thus given by (see Definition 6.1):
pO (t1) = mv IvG/O (t1) +
I
N
mj (Bvj/G(t1) + IvG/O (t1))
j =1
+ m(BvP/G(t1) + IvG/O (t1)),
where mv is the mass of the vessel in the control volume and mj is the mass of the j th
particle in the control volume. In writing the inertial velocity of m as the sum of the
inertial velocity of the control volume plus the relative velocity, BvP/G, we have used
Assumption 6.9 (the control volume is not rotating) and the results of Section 3.6.
Note that G here is not the center of mass of the entire system of particles and
thus does not obey Newton’s second law as in Eq. (6.16). Nevertheless, we use the
fact that we can write the integrated form of Newton’s second law for the total system
of particles (including the differential mass about to enter the control volume) as
in Eq. (6.10) to find an equation of motion for the control volume. We thus write
the impulse form of Newton’s second law for all particles between times t1 and
t1 + t:
t2 =
mv IvG/O (t2) +
N
mj (Bvj/G(t2) + IvG/O (t2)) + m(BvQ/G(t2) + IvG/O (t2))
j =1
= mv IvG/O (t1) +
N
mj (Bvj/G(t1) + IvG/O (t1))
j =1
(ext)
+ m(BvP/G(t1) + IvG/O (t1)) + Fcv (t1, t2),
LINEAR MOMENTUM OF A MULTIPARTICLE SYSTEM
223
where BvQ/G is the relative velocity of mass m which left the control volume at
(ext)
t2 (by Assumption 6.8) and Fcv (t1, t2) is the impulse corresponding to the external
force acting on the control volume over t.
From Assumption 6.7 we have
vj/G(t2) = Bvj/G(t1).
B
This allows us to simplify and combine terms to find
mG IvG/O (t2) − IvG/O (t1) + m IvG/O (t2) − IvG/O (t1)
= m
(ext)
vP/G(t1) − BvQ/G(t2) + Fcv (t1, t2),
B
(6.66)
where mG = mv + N
j =1 mj is the total mass of material inside the control volume.
Our final step is to divide Eq. (6.66) by t and take the limit t → 0. This allows
us to use the definitions of differentiation and integration (see Appendix A) to find
lim
t→0
Id 1 I
I
vG/O (t2) − IvG/O (t1) =
vG/O (t1) = IaG/O (t1)
t
dt
and
lim
t→0
t2
1 (ext)
1
F
(t1, t2) = lim
t→0 t
t cv
t1
(ext)
(ext)
Fcv
dt = Fcv
(t1).
We also define the mass flow rate ṁ:
lim
ṁ =
t→0
m
.
t
The mass flow rate is the rate at which the mass of the fluid enters and leaves the
control volume. It is taken to be a positive number.
Finally, the second term in Eq. (6.66) after dividing by t is zero in the limit
t → 0 because it equals limt→0 ṁIaG/O t. This leaves the final equation of
motion for the center of mass of the control volume:
(ext)
mGIaG/O = Fcv
+ ṁ
vin/G − Bvout/G ,
B
(6.67)
where we have introduced the new notation
B
vin/G =
vP/G
B
B
vout/G =
vQ/G
B
for the inflow and outflow velocities, respectively. Note that it is common practice to
put the effect of the mass flow on the right of Newton’s second law, as in Eq. (6.67),
to make it look like a force (such as the thrust in a jet engine or rocket), even though
it arises from momentum balance. As we’ll see in the next example, the mass flow
has the effect of a thrust on the mass inside the control volume.
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CHAPTER SIX
I
B
b2
e2
G
v0
θ
–mge2
O
v0
θ
b1
–Ff
e1
(a)
(b)
Figure 6.21 Fluid flowing onto a diverter at angle θ with mass flow rate ṁ. (a) Flow into
diverter. (b) Free-body diagram.
Example 6.13 Force Due to Mass Flow
This example examines the force on a barrier due to a flowing fluid, as shown in
Figure 6.21a. The diverter changes the direction of the constant flow, which has
mass flow rate ṁ, by angle θ. The flow in turn exerts a force Ff on the diverter;
or, conversely, by Newton’s third law, the diverter exerts a force −Ff on the fluid.
Figure 6.21b shows a free-body diagram of the fluid (collection of particles) inside
the control volume in contact with the diverter. Assuming a steady stream of constant
cross-sectional area, the speed of the flow into and out of the volume must be the
same for the mass to stay constant (this is true by the condition of continuity) and
thus satisfies Assumptions 6.6 and 6.7. We label this speed v0. Since we are assuming
a steady state, for which the center of mass of the fluid on the diverter is fixed in
absolute space, frame B and the inertial frame I are the same. This also means that
I the acceleration of the control volume is zero: dtd IvG/O = 0. Written in terms of
components in the inertial frame, the velocities of the inflow and outflow are
B
vin/G = v0e1
B
vout/G = v0 cos θe1 + v0 sin θe2.
These velocities are substituted into Eq. (6.67) to find the total external force on the
mass of fluid in the steady state:
Ff = ṁ(v0 − v0 cos θ)e1 − (mg + ṁv0 sin θ)e2.
6.3.2 Systems Gaining and Losing Mass
The other common situation we now examine is a control volume that is continually
ejecting or collecting particles, as in Figure 6.22. The classic example is the rocket,
which produces thrust by ejecting gas at high velocity. We study the rocket problem
in detail in Tutorial 6.3.
As for the study of a steady stream of fluid, we need to make certain simplifying
assumptions to make our particle-dynamics approach tractable:
Assumption 6.6 The particles inside the control volume have zero velocity relative
to frame B fixed to the control volume at point C.
LINEAR MOMENTUM OF A MULTIPARTICLE SYSTEM
b2
B
e2
B
(ext)
Fcv
Iv
C/O
mG + mP
I
225
j
C
Iv
C/O
mG
I
(ext)
Fcv
j
C
b1
P
Control volume
O
O
e1
e3
Bv
P/C
(a)
(b)
Figure 6.22 A variable-mass system of mass mG before and after ejecting a single particle P
of mass mP . (a) t = t1. (b) t = t2 = t1 + t.
Assumption 6.7 The control volume does not rotate relative to inertial frame I .
That is, we only study translational motion of the control volume.
Note that we relaxed the assumptions from the steady-stream example. We no
longer require that the particles be evenly distributed and thus that the center of
mass be fixed in the control volume. This allows for mass to be added or subtracted
arbitrarily. The consequence is that we will be deriving equations of motion for a
geometric point fixed to the control volume rather than for the center of mass of the
material inside the control volume; the center of mass can potentially move in the
control volume if mass is added or subtracted unevenly. We also now assume that
the particles are not moving in the control volume prior to ejection (or that they are
brought instantaneously to zero relative velocity once captured). Although this is a
strong assumption, it works quite well and simplifies problem formulation. Finally,
we have also removed the restriction that the control volume has constant mass. In
this case, we allow the total mass of particles in the control volume to increase or
decrease over time as particles are added or removed.
The procedure for finding the equation of motion of the control volume is almost
exactly the same as for the steady stream, except here we consider the motion of a
point C fixed to the control volume that is not necessarily the center of mass. We first
consider only the case of mass being ejected from the control volume. Since all of
the material inside the control volume (the vessel and the fluid particles) are traveling
together at the same velocity by Assumption 6.6, the total linear momentum prior to
ejection is
pO (t1) = mv IvC/O (t1) +
I
N
mj IvC/O (t1) + mP IvC/O (t1),
j =1
where mP is the mass of the particle being ejected. The total linear momentum after
ejection is
pO (t2) = mv IvC/O (t2) +
I
N
j =1
mj IvC/O (t2) + mP (BvP/C (t2) + IvC/O (t2)).
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CHAPTER SIX
Substituting the total linear momentum at t1 and t2 into the impulse form of Newton’s
second law for a collection of particles (Eq. (6.10)) gives
mGIvC/O (t2) + mP BvP/C (t2) + mP (IvC/O (t2) − IvC/O (t1))
(ext)
= mGIvC/O (t1) + Fcv (t1, t2),
where, as before, mG = mv + N
j =1 mj . Once again, we divide by t and take the
limit t → 0. The second-order term again goes to zero, and we are left with the
equation of motion:
(ext)
mGIaC/O = Fcv
− ṁout Bvout/C ,
(6.68)
B
where, as before, we used the notation Bvout/C =
vP/C and we introduced ṁout =
limt→0 mP /t.
Recall in Chapter 2 when we emphasized that Newton’s law does not contain an
ṁP term? That is still true. The ṁout here refers to the rate that mass is ejected from the
collection. It only arises in the study of systems of particles. The rate of change of mass
term appeared because we were careful to keep track of the total linear momentum of
all particles during the time interval—a requirement of the impulse form of Newton’s
second law and the law of conservation of total linear momentum.
What about systems collecting mass? Following the same procedure, except now
considering a particle Q entering the control volume, leads to the same equation of
motion for the control volume but with the opposite sign on the mass flow. (Deriving
this result is a useful exercise.) In fact, in some cases there is mass both entering
and leaving the system (such as in a jet engine) but with different velocities. In that
case, we just include two thrust terms in Eq. (6.68). Using the convention that the
scalar inflow and outflow mass rates are both positive (ṁout , ṁin > 0), we can rewrite
Eq. (6.68) for both inflow and outflow as
(ext)
mGIaC/O = Fcv
− ṁout Bvout/C + ṁin Bvin/C ,
(6.69)
where ṁin is the mass inflow rate and Bvin/C is the inflow velocity. Note that both flow
velocities are relative to frame B fixed to point C of the control volume. Also note
that the mass mG of the collection varies with time according to ṁG = −ṁout + ṁin .
Example 6.14 Loading a Dump Truck
Consider a dump truck of unloaded mass M = 2,000 kg being filled from a chute that
is at angle θ = 30◦ from the horizontal, as shown in Figure 6.23a. The material leaves
the chute with a speed of u = 2 m/s and a mass flow rate of 75 kg/s. Suppose that the
driver forgot to put on the parking brake and left the truck in neutral; the truck will
thus accelerate as material continues to be added. We will calculate the speed of the
truck as a function of time.
LINEAR MOMENTUM OF A MULTIPARTICLE SYSTEM
227
I
B
b2
u
30°
O
Nb2
v
C
Control volume
C
b1
–mGgb2
(a)
(b)
Figure 6.23 (a) Material falling into a dump truck from a chute at angle θ. (b) An attached
control volume and reference frames.
Figure 6.23b shows the frames of reference and control volume surrounding the
truck. The truck is constrained to move only in the b1 direction, so its velocity is
Iv
C/O = vb1. The inflow velocity of material is given by
vin/C = Ivin/C = Ivin/O − IvC/O = (u cos θ − v)b1 − u sin θb2.
B
Using IaC/O = v̇b1, we can substitute into Eq. (6.69):
(M + ṁin t)v̇b1 = (N − (M + ṁin t)g)b2 + ṁin ((u cos θ − v)b1 − u sin θb2),
where we have used the fact that mG = M + ṁin t, the total mass in the control volume,
(ext) = (N − (M + ṁ t)g)b (we are ignoring any frictional forces on the tires
and Fcv
in
2
and assuming only a vertical normal force). Considering the b1 components gives the
differential equation of motion for the truck:
v̇ =
ṁin (u cos θ − v)
.
M + ṁin t
This equation is integrated using v̇ = dv/dt and rearranging terms:
v(t)
0
1
dv =
u cos θ − v
t
0
ṁin
ds.
M + ṁin s
Completing both integrals yields
ln
M + ṁin t
u cos θ
.
= ln
M
u cos θ − v(t)
This equation can be solved for v(t):
v(t) = u cos θ −
M
M + ṁin t
u cos θ =
ṁin tu cos θ
.
M + ṁin t
228
CHAPTER SIX
For instance, the speed of the truck after 2 s is
v(2) = 0.12 m/s.
In the problems you have the opportunity to find the final speed of the truck when it
passes the chute.
6.4 Tutorials
Tutorial 6.1 Two Falling Masses Connected by a Spring
This tutorial considers the simple model of two particles falling vertically under the
influence of gravity, the lower one of mass M and the upper one of mass m < M. This
model illustrates a situation when the motion of the center of mass cannot be treated
independently of the motion relative to the center of mass. The particles are connected
by a spring with spring constant k and unstretched length l, as shown in Figure 6.24.
Since the particles are falling in the ey direction only, each has one degree of freedom
and the problem thus has two degrees of freedom, represented by the coordinates yM
(the height of the lower mass) and ym (the height of the upper mass). We consider
two cases: (a) the gravitational force acting on the two masses is a constant (Mg for
the lower mass and mg for the upper one) and (b) the gravitational force is the more
realistic 1/r 2 force (see Example 4.9). These cases are shown in the two free-body
diagrams, Figure 6.24b and c.
From Eq. (6.13) we find the center-of-mass position rG/O of the two particles,
rG/O =
ym
I
ey
m
k
yM
1
y G ey .
MyM + mym ey =
M +m
g
–mgey
M
–Tey
μm
– —2 ey
(RE + ym)
Tey
O
–Mgey
ex
(a) Coordinates and reference frame
(b) Free-body diagram 1
–Tey
Tey
μM
– —2 ey
(RE + yM)
(c) Free-body diagram 2
Figure 6.24 (a) Two masses connected by a spring falling vertically under two different gravity
models. (b) Constant-gravity model. (c) Model with 1/r 2 gravity.
LINEAR MOMENTUM OF A MULTIPARTICLE SYSTEM
229
Consider for a moment the case where the masses are rigidly connected by a rod
rather than by a spring. Then the position of the center of mass is fixed relative to the
two particles (a function of yG and l only). For the constant-gravity case, the equation
of motion for the center of mass is given by Eq. (6.16):
ÿG = −g.
For the more realistic gravity case, we use Figure 6.24c and ym − yM = l to find
⎤
⎡
ÿG =
−1 ⎢
μM
⎣
M +m
RE + y G −
μm
2 + m
RE + y G +
m+M l
M
m+M l
⎥
2 ⎦ , (6.70)
where μ = GmE , and mE and RE are the mass and radius of the earth, respectively.
In both cases, the equations of motion are in terms of the center-of-mass position
only; it doesn’t matter that the forces act on each particle away from the center of
mass. As expected, the two rigidly connected particles behave as a single particle of
mass m + M.
Nevertheless, in the more realistic gravity case, the force is more complicated than
just that on a single particle. It is possible to find another point between the two masses
where the force of gravity of a virtual particle of mass M + m would be the same as in
Eq. (6.70). If we call that point Q, its location can be found for this simple geometry
by solving for yQ from
μM
μ(M + m)
=
2
(RE + yQ)
RE + y G −
μm
2 + m
RE + y G +
m+M l
M
m+M l
2 .
The point yQ is where the force of gravity on an effective particle of the same total
mass is equal to the net force of gravity on all the particles; we call it the center of
gravity. We could formulate the translational equation of motion in terms of the center
of gravity, but, as shown in the next chapter, we would no longer have separation of
translation and rotation. For a general collection of particles (or a rigid body), the
center of gravity is not the same as the center of mass. It is a common error to refer to
the center of gravity instead of the center of mass when discussing the translational
motion of a collection of particles.
The situation is different when the masses are connected by a spring. For the case
of a constant gravity field, we can use the free-body diagram in Figure 6.24b to write
the equations of motion for each mass:
k
k
ym − yM = −g +
m
m
k
k
ÿM + yM − ym = −g −
M
M
kl
m
kl
,
M
where the spring force in Figure 6.24b is given by T = −k l − (ym − yM ) .
We can use the definition of the center of mass to take the weighted sum of these
equations and find, as expected, that the internal spring force cancels out and the
ÿm +
230
CHAPTER SIX
equation of motion for the center of mass is the same as for the rigid connection:
ÿG = −g.
Again, the center of mass behaves like a single particle acted on by the total external
gravity force. In this case, the center-of-mass motion separates completely from
the oscillatory motion of the two particles about the center of mass (which, for the
constant-gravity case, is zero if the masses start a distance l apart).
For the more realistic gravity force, the two equations of motion are
ÿm +
ÿM +
k
μ
k
kl
ym − yM +
=
m
(RE + ym)2
m
m
k
μ
k
kl
yM − ym +
=− .
2
M
M
(RE + yM )
M
These equations of motion are slightly more complicated. Taking the weighted sum
yields the center-of-mass equation of motion,
μ
m
M
ÿG +
+
= 0.
(6.71)
m + M (RE + ym)2 (RE + yM )2
The internal spring force still cancels out (as expected from Eq. (6.16)). However,
for the more complicated gravity force, the center-of-mass equation of motion, while
still a function of the total external force, depends on the position of the individual
particles. Eq. (6.70) is the same as Eq. (6.71) only when ym − yM = l. There is no
way to make the center-of-mass equation depend only on yG. This is an example
of when the external force introduces a dependence on the relative position of the
particles. Eq. (6.16) does not imply that the center-of-mass equation always separates
completely from the individual particle motions.
It is interesting to note that formulating the equation of motion for the center
of mass looks like we added an additional degree of freedom (that of a “virtual”
particle located at the center of mass), leaving us with three equations rather than
two. However, at the outset of this example we noted that it was a two-degree-offreedom problem. It still is; the three equations represent two degrees of freedom
with the added constraint from the center-of-mass corollary in Eq. (6.15).
Tutorial 6.2 Bouncing Ball
This tutorial studies the dynamics of a falling ball that bounces off a rigid, fixed
inclined plane. The flight of the ball has three phases: before, during, and after the
bounce. We solve for the trajectory of the ball in each phase and find the bounce
distance as a function of the inclined plane angle. We also numerically compute the
plane inclination angle that maximizes the bounce distance.
Let particle P represent the ball. We use the Cartesian coordinates xP and yP
to describe the position of P , as shown in Figure 6.25a. The inertial frame I =
(O, ex , ey , ez ) and the collision frame C = (O , en, et , e3) are shown in Figure 6.25b,
where O = O . The collision frame and the inertial frame are related by the transformation
LINEAR MOMENTUM OF A MULTIPARTICLE SYSTEM
yP
I
ey
g
P
231
C
y0
FPen
en
xP
P
O = O′
O
ex
θ
–mPgey
et
(a) Coordinates
(b) Reference frames
(c) Impact free-body diagram
Figure 6.25 Bouncing ball. During freefall, the only force acting on P is gravity.
et
en
ex
ey
cos θ
sin θ
− sin θ
cos θ ,
(6.72)
where we assume 0 < θ < π/4.9
We write the velocity of P in frame C as IvP/O = uP et + vP en. Assume P starts
falling (from rest) at t = 0. Let t1 and t2 denote the times immediately before and after
t2 − t1 is very short. Also assume that we have previously
the collision, where t =
conducted an experiment to estimate the coefficient of restitution e for the impact of
the ball and the plane (see Example 6.12).
As in Example 6.12, we use energy to find that the initial velocity before impact is
I
vP/O (t1) = − 2gy0ey .
(6.73)
Eq. (6.73) expressed as components in the collision frame is
I
vP/O (t1) = 2gy0(sin θet − cos θen).
(6.74)
From Example 6.12 we know that Eq. (6.60) is a valid approximation of the coefficient
of restitution e for the collision between a particle and a rigid fixed surface (with the
error being proportional to t 1). We thus drop the O(t) terms to find
(6.75)
vP (t2) = −evP (t1) = e 2gy0 cos θ.
But what about the tangential speed uP (t2) after impact? Since the plane is held
fixed by external forces, we cannot use conservation of total linear momentum.
9 The
upper bound on θ is a consequence of the following observation: if the coefficient of restitution
satisfies e = 1 and θ = π/4, then the ball bounces in the ex direction. Hence if e < 1 and θ = π/4, the ball
bounces in the ex and −ey direction (to the right and down). Since we would like to measure the horizontal
bounce distance at the height of the impact, we need the upper bound θ < π/4. The lower bound ensures
the ball bounces to the right.
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CHAPTER SIX
Nonetheless, we know that the impulse FP (t1, t2) acts only in the normal direction,
which implies that the only tangential force on P during the collision is gravity. Using
the impulse form of Newton’s second law, we observe
t2
I
vP/O (t2) = IvP/O (t1) + FP (t1, t2) +
(−mP gey )
t1
= IvP/O (t1) + F P (t1, t2) − mP g cos θt en + mP g sin θtet ,
which implies
uP (t2) = uP (t1) + mP g sin θt .
(6.76)
=O(t)
At impact the tangential speed of P increases by O(t). In the limit t → 0, the
tangential speed of P is conserved. Thus we take uP (t2) = uP (t1) with a small error
on the same order as in the normal direction. Using Eqs. (6.74)–(6.76), we obtain the
velocity of P after the impact:
I
vP/O (t2) =
2gy0 sin θ et + e 2gy0 cos θ en
= (1 + e) 2gy0 sin θ cos θex +
2gy0(e cos2 θ − sin2 θ)ey .
(6.77)
We now compute the trajectory of P during the final phase of flight by integrating
the equation of motion
−mP gey = mP IaP/O
from t2 to t > t2 using the initial conditions IvP/O (t2) given by Eq. (6.77):
xP (t) = ẋP (t2)(t − t2)
(6.78)
1
yP (t) = − g(t − t2)2 + ẏP (t2)(t − t2).
2
(6.79)
At t = tf , which is the time when P lands after the bounce, yP (tf ) = 0 and
Eqs. (6.78) and (6.79) yield
xP (tf ) =
2ẋP (t2)ẏP (t2)
g
= 2(1 + e)y0 sin 2θ(e cos2 θ − sin2 θ).
For y0 = 1 m, Figure 6.26 shows xP (tf ) as a function of the plane incline angle θ
and several values of e. Note that the bounce distance decreases to zero as the coefficient of restitution decreases to zero. We also plot the optimal angle that maximizes
xP as a function of e. For example, if e = 1, then the optimal angle is θ ≈ 22.5◦, which
corresponds to an initial flight angle of 45◦ relative to ex . Notice that the optimal angle
also decreases to zero as the coefficient of restitution decreases to zero.
LINEAR MOMENTUM OF A MULTIPARTICLE SYSTEM
233
2.0
e = 0.25
e = 0.50
e = 0.75
e = 1.00
Optimal θ
xP for y0 = 1 (m)
1.6
1.2
0.8
0.4
0.0
0
10
20
30
40
θ (°)
Figure 6.26 Bouncing ball. The optimal incline angle maximizes the bounce distance xP for
a given value of the coefficient of restitution e.
Tutorial 6.3 The Rocket Problem
A common example of a variable-mass system is a rocket. A rocket expels mass
to provide thrust, which accelerates the rocket. Figure 6.27 shows the appropriate
frames and a free-body diagram of a rocket. The inertial frame I = (O, e1, e2, e3) is
fixed to the ground with e1 pointing upward along the path of the rocket. The frame
B = (G, ex , ey , ez ) is fixed to the rocket at its center of mass G and aligned with the
inertial frame. The inertial velocity of the rocket is
I
vG/O = vGex .
The exhaust velocity of the gas is Bvout/G = −ue ex . Using Figure 6.27 and Eq. (6.69)
in the ex direction, with C replaced by G, we find
mGv̇G = Ae P − mGg + ṁout ue ,
(6.80)
where Ae P is the external force on the rocket due to the difference between
atmospheric pressure and the pressure of the exhaust gas right at the nozzle opening
(Ae is the exit area of the nozzle).
Eq. (6.80) is known as the rocket equation.It applies to all propulsion systems, from
the small jets used to control a satellite to the enormous chemical engines used to loft
the Saturn V to the moon.10 The primary design parameters available to the engineer
are the mass (out)flow rate ṁout and the exhaust velocity ue . For large chemical
rockets, the mass flow rate is usually constant and determined by the pressure in
10 A chemical rocket is an engine that employs combustion among volatile reactants and oxygen to heat
up the resulting gas and accelerate it out through the nozzle.
234
CHAPTER SIX
ex
vG ex
g
AeΔPex
ey
B
G
G
I
G
–mGgex
e1
O
e2
–ueex
(a) Reference frames
(b) Kinematics
(c) Free-body diagram
Figure 6.27 The rocket problem.
the fuel tank, so the exhaust velocity is the main driver. This can be seen by solving
Eq. (6.80) using ṁout = −ṁG (the mass loss of the rocket due to the outflow of burning
fuel). For large engines with substantial thrust, the pressure term is usually very small
and can be ignored. Thus, to find the final velocity of the rocket, we need to solve
dvG
1 dmG
= −ue
− g,
mG dt
dt
(6.81)
which is a separable differential equation (see Appendix C). Integrating Eq. (6.81)
from t0 to tf yields
v = ue ln
m0
− g(tf − t0),
mf
mG(t0) is the
vG(tf ) − vG(t0) is the change in rocket velocity, m0 =
where v =
initial mass of the rocket, and mf = mG(tf ) is the final mass of the rocket.
For a constant mass flow rate, the solution can be written in terms of the change
in mass of the rocket only by substituting m0 − mf = ṁout (tf − t0):
v = ue ln
m0
g
−
(m0 − mf ).
mf
ṁout
(6.82)
Note that this solution is an approximation for rockets launched from the surface of
the earth, as we have assumed that the acceleration g due to gravity is a constant.
The second term in Eq. (6.82), however, is normally small compared to the first (and
essentially zero for deep-space propulsion), so we can ignore it and still get reasonable
LINEAR MOMENTUM OF A MULTIPARTICLE SYSTEM
235
estimates for rocket performance. Ignoring this term and inverting Eq. (6.82) yields
the rocket equation of Tsiolkovsky,11
mf
m0
= e−v/ue .
(6.83)
Eq. (6.83) is an extremely useful relationship between the fraction of mass brought
to altitude as a function of the needed velocity change and the exhaust velocity of the
rocket. Recall from Tutorial 4.3 that we typically treat orbital maneuvers by means
of impulsive changes in velocity. Eq. (6.83) indicates how much fuel is needed for a
given v, assuming a specific type of rocket characterized by ue .
Now we can plug in some numbers for launch from the surface of the earth to orbit.
The best large chemical rockets have an exhaust velocity in the range of 2–3 km/sec.
Getting from the surface of the earth to low-earth orbit requires a velocity increment
on the order of 6–7 km/sec. Thus a single rocket engine going all the way to orbit
would have a final payload mass fraction of (best case) e−2 ≈ 0.15. That means that
at least 85% of the rocket mass is fuel! Add in all structural mass and fuel tanks, and
very little is left for payload. It is extremely difficult to launch satellites into space
on a single rocket, usually referred to as a single-stage-to-orbit vehicle. That is why
most rockets today have multiple stages.
6.5 Key Ideas
.
The total linear momentum of a collection of particles is the sum of the individual
inertial momenta:
I
pO =
N
mj Ivj/O .
j =1
.
The total linear momentum of a collection of particles satisfies Newton’s second
law, which in integrated form is
I
pO (t2) = IpO (t1) + F
where F
.
(ext)
(ext)
,
is the total external impulse.
The law of conservation of total linear momentum states that, if the total external
force is zero, the total inertial linear momentum of a system of particles is a constant
of the motion:
N
j =1
11 Konstantin
mj Ivj/O (t2) =
N
mj Ivj/O (t1).
j =1
Tsiolkovsky (1857–1935) was a pioneering Russian rocket scientist. He was among the first
to develop the theory of rocketry, publishing the famous “rocket equation” in 1903. Often considered
the father of modern rocketry, he was visionary and prescient in both his fiction and nonfiction writings
regarding satellites and manned space travel.
236
CHAPTER SIX
.
The center of mass G of a collection of particles is the vector sum of the particle’s
mass-weighted positions, normalized by the total mass mG = N
j =1 mj :
rG/O =
.
N
1 mi ri/O .
mG i=1
The center-of-mass corollary states that the vector sum of the mass-weighted
positions of each particle relative to the center of mass is zero:
N
mi ri/G = 0.
i=1
.
The center of mass obeys an equation of motion corresponding to an equivalent
particle of mass mG located at the center of mass,
Id
(ext)
pG/O = FG
.
I
dt
In the absence of external forces, the center of mass is either stationary or moving
at a constant inertial velocity.
.
The motion of a particle i in a multiparticle system relative to a frame B fixed to
the center of mass is
mi Bai/G = Fi(ext) +
N
Fi,j − mi IaG/O ,
j =1
where Fi,j is the internal force on particle i due to particle j .
.
In a collision between two particles, P and Q, we find the velocities after the
collision (t ≥ t2) using the velocities before the collision (t ≤ t1) expressed in a
collision frame C = (O , en, et , e3). The origin of the collision frame is the point
of impact and the normal direction is
en = Iv̂P/Q(t1).
In general the tangential components of the velocities are constant and the normal
components obey the law of conservation of total linear momentum,
mP vP (t2) + mQvQ(t2) = mP vP (t1) + mQvQ(t1).
The normal components also obey the coefficient-of-restitution equation,
e=
.
vP (t2) − vQ(t2)
vQ(t1) − vP (t1)
.
In a collision between a particle and a fixed surface, the tangential components
are constant, but the total linear momentum is no longer conserved for the normal
components. The coefficient-of-restitution equation is
e=−
vP (t2)
.
vP (t1)
LINEAR MOMENTUM OF A MULTIPARTICLE SYSTEM
.
237
For a variable-mass system with mass flowing in a steady stream at a rate ṁ, the
equation of motion for the center of mass is
(ext)
mGIaG/O = Fcv + ṁ Bvin/G − Bvout/G ,
where Bvin/G and Bvout/G are the inflow and outflow velocities.
.
For a variable-mass system with mass flowing in at a rate ṁin and out at a rate
ṁout , the equation of motion for the center of mass is
mG(t)IaG/O = FG − ṁout Bvout/G + ṁin Bvin/G,
where Bvout/G is the relative velocity of the outflowing mass and Bvin/G is the
relative velocity of the inflowing mass.
6.6 Notes and Further Reading
Although Newton formulated the basic laws of mechanics, it was Euler who showed
they could be applied to collections of particles independently and first developed the
techniques of multiparticle systems we presented here. (He also defined the center
of mass and formulated the N -body problem.) In modern texts it forms the basis of
the development of rigid-body motion, which we discuss in Chapter 9. Almost all
introductory texts have some discussion of multiparticle systems and impacts.
Most introductory texts (e.g., Greenwood 1988; Pytel and Kiusalaas 1999; Meriam
and Kraige 2001; Bedford and Fowler 2002; Hibbeler 2003; Tongue and Sheppard
2005; Rao 2006; Beer et al. 2007) also introduce simple models of mass flow or
variable mass systems, though the level of detail varies widely. In our treatment we
have tried to be explicit about the assumptions being made and the problems to which
they apply. A more careful treatment requires the development of fluid mechanics, of
which this is only the briefest introduction. An excellent starting text is Smits (1999).
Versions of Example 6.13 appear in almost every text; ours is most similar to that in
Bedford and Fowler (2002). Likewise, Example 6.14 is similar to problems in Pytel
and Kiusalaas (1999), Bedford and Fowler (2002), Tongue and Sheppard (2005), and
Beer et al. (2007).
6.7 Problems
6.1
Consider four masses whose Cartesian coordinates (x, y)I are given in Figure 6.28. Find the position of the center of mass.
(1, 2)I
I
2m
(2, 1)I
(0, 1)I
m
5m
m
O
(1, 0)I
Figure 6.28 Problem 6.1.
238
CHAPTER SIX
M
g
m
Figure 6.29 Problem 6.2.
6.2
Two masses are connected by a string, as shown in Figure 6.29. The larger
mass M slides on a surface with coefficient of friction μ. Find the equation
of motion for each mass.
6.3
Consider a double mass-spring system with two masses of M and m on a
frictionless surface, as shown in Figure 6.30. Mass m is connected to M by a
spring of constant k and rest length l0. Mass M is connected to a fixed wall by
a spring of constant k and rest length l0 and a damper with constant b. Find
the equations of motion of each mass. [HINT: See Tutorial 2.1.]
I
b
M
k
m
k
O
Figure 6.30 Problem 6.3.
6.4
Use matlab to find and plot the solution to the equations of motion in
Problem 6.3, assuming M = 10 kg, m = 2 kg, k = 0.5 N/m, b = 0.2 N-s/m,
l0 = 0.5 m, and each mass starts from rest. Set the initial distance between
mass M and the wall to be l0/2 and the initial distance between mass m and
M to be l0. Be sure to label your axes with units and add a legend.
6.5
Masses P and Q collide and stick together to form mass PQ, as shown
in Figure 6.31. Assuming mass P weighs twice as much as mass Q, find the
I
30°
5 m/s
P
4 m/s
Q
O
Figure 6.31 Problem 6.5.
LINEAR MOMENTUM OF A MULTIPARTICLE SYSTEM
239
speed and orientation of the velocity of PQ, measured relative to the initial
heading of mass Q.
6.6
Three seconds remain in the fourth quarter of the Super Bowl, and the offense
is down by four points (Figure 6.32). On fourth and goal, the wide-receiver
lines up on the 15 yard line. Sensing a defensive blitz, the quarterback calls an
audible that directs the receiver to sprint directly down the sideline toward the
endzone. If the receiver runs at 8 yd/s and the quarterback immediately throws
the ball at 12 yd/s, at what angle θ should the quarterback aim to connect with
his receiver? If the ball is caught, will he catch the ball in the endzone?
10 yds
15 yds
vP
P
θ
vQ
O
25 yds
Figure 6.32 Problem 6.6.
6.7
Show that the linear momentum of the center of mass IpG/O of a collection
of N particles is equal to the total linear momentum of the collection, IpO =
N
I
j =1 mi vi/O . Please state any assumptions that you make.
6.8
Consider a model of two cars colliding, as in Figure 6.33. Car B was traveling
south with speed vB and car A was traveling northeast at angle θ1 and at
speed vA when they collided. After the crash, the two cars stuck together and
skidded off at angle θ2. Each driver claimed to be traveling at the speed limit
of 30 mph (48 km/hr) and each claimed to have slowed down but collided
anyway because the other car was speeding. Assume the mass of car A is mA
and the mass of car B is mB .
a. Assume that mB /mA = 0.9. If θ1 = 40◦ and θ2 = 15◦, which car was
going faster?
North
θ2 vf
vA
θ1
A
B
vB
Figure 6.33 Problem 6.8.
240
CHAPTER SIX
b. What was the speed of the faster of the two cars if the slower car was
indeed going at the speed limit?
6.9
Masses P and Q are attached by a massless rigid rod to form a dumbbell, as
shown in Figure 6.34. The dumbbell is tossed in the presence of gravity so
that the masses may spin about the center of mass G. Show that the trajectory
of the center of mass is a parabola.
v0
P
g
G
Q
Figure 6.34 Problem 6.9.
6.10 An Olympic biathlete must fire a rifle at a target while standing on crosscountry skis, as shown in Figure 6.35. Suppose a bullet of mass m is fired
with speed v0, and the biathlete and rifle have a combined mass of M.
a. If the ski/snow interface is frictionless, how fast does the biathlete slide
backward after firing at the target?
b. Now suppose that there is friction between the skis and the snow. If
the bullet leaves the rifle barrel in a time interval of t after the trigger
is pulled, what is the minimum coefficient of friction that will prevent
the biathlete from sliding?
Figure 6.35 Problem 6.10.
6.11 A planar, quarter-car model of an automobile suspension system is shown
in Figure 6.36. The model consists of two masses: mass M represents onequarter of the car and mass m represents the tire. The masses M and m are
connected by a spring with constant k1 and rest length l1 and a damper with
constant b. Mass m is connected to the ground by a spring with constant k2
and rest length l2. Find the equations of motion for x1 and x2. [HINT: The
damper force on M is −bẋ1e2.]
LINEAR MOMENTUM OF A MULTIPARTICLE SYSTEM
241
M
g
k1, l1
b
x1
m
e2
x2
k2, l2
e1
O
Figure 6.36 Problem 6.11.
6.12 An inverted simple pendulum of mass m and length l is attached to a cart
of mass M that slides without friction on a horizontal track subject to an
unspecified force F (Figure 6.37). Find the equations of motion of x and θ.
m
l
g
I
θ
x
F
M
O
Figure 6.37 Problem 6.12.
6.13 Figure 6.38 shows two balls falling, a larger bottom ball P and a smaller top
ball Q, just resting on top of the bigger ball (e.g., a tennis ball and a basketball).
The lower ball has mass mP and the upper one has mass mQ. Assume that e1
is the coefficient of restitution between the larger ball and the ground and e2
is the coefficient of restitution between the smaller and larger balls. Suppose
that they fall from height h, such that their joint speed just before the lower
ball hits the ground is v0.
vQ
Q
Q
v0
P
P
(a)
vP
(b)
Figure 6.38 Problem 6.13.
242
CHAPTER SIX
a. Find an expression for the upward speed of ball Q after the collision
with the ground in terms of e1, e2, mP , mQ, and v0.
b. Show that for mP
mQ and e1 ≈ e2 ≈ 1 (a perfectly elastic collision),
the top ball triples its speed, vQ ≈ 3v0. (This is a fun experiment to do,
but do it outside!)
6.14 Consider a model for two stationary rail cars, B and C, of masses mB and
mC , connected together by a spring coupling of spring constant k. A third car
A, of mass mA, is traveling along the frictionless track at a constant speed vA
until it collides with the two coupled cars. Model the three rail cars as point
masses, arranged as in Figure 6.39. Assume that the coupling is broken on car
A and that the collision of car A with car B is perfectly elastic (e = 1). What
is the speed of the center of mass of cars B and C after the collision?
k
vA
A
B
C
Figure 6.39 Problem 6.14.
6.15 Figure 6.40 shows two masses on springs arranged to collide with each other
when the springs are at their unstretched lengths l0. At time t = 0, both springs
are compressed an identical amount x0. Assuming each mass has identical
mass m, each spring has the same spring constant k, and the coefficient of
restitution between the masses is e, find an expression for the maximum
spring compression after the nth collision. Using matlab, plot maximum
spring compression versus collision number for the first 10 collisions. Use
the following parameters: x0 = 10 m, l0 = 5 m, e = 0.9, 0.75, 0.5, and 0.2.
Plot the results on a single graph for all values of e (so there should be four
plots on the graph).
x
l0
Figure 6.40 Problem 6.15.
6.16 Two small balls, each of mass m = 0.2 kg, hang from strings of length L = 1.5
m. The left ball is released from rest with θ = −45◦. As a result of the initial
collision the right ball swings through a maximum angle of 30◦. Determine
the coefficient of restitution between the two balls.
6.17 A jet of water of cross-section A and density ρ moves horizontally at inertial
speed v0 and hits a block of mass M, which is connected to a wall with a
spring of constant k, as in Figure 6.41. Assume that the collision is inelastic
(i.e., the water leaves with a zero horizontal component of velocity relative
to the block) and that the spring is initially at its rest length l0. Find the final
length of the spring.
LINEAR MOMENTUM OF A MULTIPARTICLE SYSTEM
243
M
v0
A
k, l0
Figure 6.41 Problem 6.17.
6.18 A jet of water of cross-section A and density ρ moves horizontally at an inertial
velocity v0 and hits a block of mass M, as in Figure 6.42. Assume that the
collision is inelastic (i.e., the water leaves with a zero horizontal component
of velocity relative to the block) and that there is a coefficient of friction μ
between the block and the surface on which it slides. Find the final speed vf
of the block.
M
vf
v0
A
μ
Figure 6.42 Problem 6.18.
6.19 A nozzle with exit area A emits a column of air of density ρ, which lifts a ball
of mass M, as shown in Figure 6.43. If the inertial speed of the air column is
v0, find the height h to which the ball is raised.
g
v0
h
A
Figure 6.43 Problem 6.19.
6.20 A turbojet (shown in Figure 6.44) uses a steady stream of air to accelerate
an airplane. It operates by taking in air via a compressor and accelerating
it by means of a combustion and compression process. The air is assumed
244
CHAPTER SIX
m
vG
vout
G
Figure 6.44 Problem 6.20.
stationary in the inertial frame and the mass flow rate ṁ is largely independent
of the speed of the plane. Assuming that the jet accelerates the air to a speed
vrel relative to the jet’s center of mass G, draw a control volume and find an
expression for the thrust produced by the engine.
6.21 Suppose a jet engine takes in air at a mass flow rate of 100 kg/s. Also assume
that the exhaust velocity of the air is 500 m/s relative to the plane. Determine
the magnitude of the thrust on the plane for two different cruising speeds:
(a) 460 km/h and (b) 980 km/h.
6.22 Suppose the container on the dump truck in Example 6.14 has a length of 5 m.
What is the final speed of the dump truck when it clears the chute?
6.23 A “kinetic kill” missile destroys its target through conversion of kinetic energy
to mechanical work. That is, there is no explosion—it destroys the target only
by means of the collision. Consider a missile with a rocket engine flying
a straight-line path to a target. Suppose the engine has a mass flow rate of
0.5 kg/sec and an exhaust velocity of 1 km/s relative to the rocket. Assuming
that the missile body has a mass of 1,000 kg, find the total mass of fuel
necessary to achieve a final kinetic energy of 5 × 106 J.
6.24 A thrust test stand is used to directly measure the thrust of a rocket. One simple
arrangement is to attach the rocket engine to a spring/damper system and
measure the spring compression over time, as shown in Figure 6.45. Consider
a rocket engine of dry mass of 30 kg loaded with 10 kg of fuel. It has a mass
flow rate of 0.5 kg/s and an exhaust velocity ue = 20 m/s. It is attached to a
spring of spring constant k = 50 N/m and rest length l0 = 0, and to a damper
with constant b = 5 N.s/m. Find the equations of motion and, using matlab,
plot the rocket’s position x(t) as a function of time. Integrate for the amount
of time that there is any fuel remaining in the rocket. How might you use this
measurement to find the thrust T = ṁout ue ?
b
m
ue
M
k, l0
x
Figure 6.45 Problem 6.24.
6.25 How high will your model rocket fly? Assume the model rocket weighs 5 kg
and the fuel weighs 0.5 kg. Assume a mass flow rate of 0.1 kg/s and an exhaust
velocity of 15 m/s. What is the final altitude of the rocket when all the fuel is
consumed? How good was the constant-gravity assumption?
CHAPTER SEVEN
Angular Momentum and Energy of a
Multiparticle System
This chapter continues our study of multiparticle systems by examining the total
angular momentum and energy of the collection. Recall the following important result
from Chapter 6: we can treat a collection of particles as a single equivalent particle
located at the center of mass, whose translational motion obeys Newton’s second law.
If the external forces on the collection are independent of the position of the particles
relative to the center of mass, then the motion of the center of mass is independent of
the relative motion of the particles. This result is used in combination with equations
describing the motion of the particles relative to the center of mass, in terms of either
linear or angular momentum, to find the entire trajectory.
In this chapter we take a slightly different approach to the motion of the particles
about the center of mass. By studying the total angular momentum and total energy of
the collection, we obtain results that not only make some problems easier to analyze,
but also open the door to the study of rigid bodies in Parts Three and Four. The most
important discovery in this chapter is that the angular momentum of the collection
dynamically separates; that is, we can treat the angular momentum of the center of
mass of the system (relative to the origin of the inertial frame) independently from
the motion of the particles about the center of mass. This result is profound and is
critical to our later treatment of rigid bodies.
7.1 Angular Momentum of a System of Particles
Consider again a system of N particles, as shown in Figure 7.1, in which each particle
N
i = 1, . . . , N , is subject to external force Fi(ext) and internal force Fi(int) =
j =1 Fi,j ,
where Fii = 0. Recall that the internal forces Fi,j are equal and opposite, that is,
246
CHAPTER SEVEN
I
i
e2
Fi,j
ri/G
ri/O
Fj,i
G
j
Fj(ext)
rG/O
O
e1
e3
Figure 7.1 Collection of particles with center of mass G.
Fi,j = −Fj,i , by Newton’s third law. The angular momentum of particle i relative to
the origin O is then, by Definition 4.2,
hi/O = mi ri/O × Ivi/O .
I
As done in Chapter 6 for linear momentum, we can write Newton’s second law in
angular momentum form separately for each particle,
Id
(int)
(ext)
hi/O = Mi/O
+ Mi/O
,
I
dt
(ext)
= ri/O × Fi(ext) is the moment on particle i due to the total external force
where Mi/O
(int)
= ri/O × Fi(int) is the moment on particle i due to the total internal
on i, and Mi/O
force on i from the other N − 1 particles in the collection. We could thus compute the
trajectories of the N particles by solving these N vector equations of motion instead
of using Eq. (6.1). Although there are certainly problems that can be solved this way,
this approach does not really provide any new tools for modeling the behavior of a
multiparticle system. Instead, we introduce the total angular momentum, just as we
did for the total linear momentum in Section 6.1.2.
7.1.1 The Total Angular Momentum
The definition of total angular momentum is similar to that of total linear momentum.
Definition 7.1 The total angular momentum of a system of particles about O is
the sum of the individual angular momenta:
I
hO =
N
i=1
I
hi/O =
N
i=1
mi ri/O × Ivi/O .
ANGULAR MOMENTUM AND ENERGY OF A MULTIPARTICLE SYSTEM
247
The utility of the total angular momentum becomes apparent when we examine its
inertial derivative. Taking the derivative of IhO and using the product rule gives
Id
dt
I
hO =
N
I
I
mi vi/O × vi/O +
i=1
=0
N
mi ri/O ×
i=1
Id
dt
vi/O .
I
(7.1)
As with linear momentum, the second term can be rewritten using Newton’s second
law, since
N
Id I
vi/O = Fi(ext) +
Fi,j .
mi
dt
j =1
This results in the following expression for the derivative of the total angular momentum,
Id
dt
I
hO =
N
i=1
ri/O × (Fi(ext)
+
N
Fi,j ).
(7.2)
j =1
We now examine the second term in the sum. Since the order in which the
forces are summed is arbitrary, it must be true that i j ri/O × Fi,j = j i rj/O
×Fj,i (we just swapped the labels i and j ). Thus
N
N ri/O × Fi,j
i=1 j =1
N N
1 =
(ri/O × Fi,j + rj/O × Fj,i ).
2 i=1 j =1
Again invoking Newton’s third law, as we did with linear momentum, and making
the substitution Fi,j = −Fj,i yields
N
N i=1 j =1
ri/O × Fi,j =
N N
1 (ri/O − rj/O ) ×Fi,j .
2 i=1 j =1 (7.3)
= ri/j
In Chapter 6 we were able to eliminate the double sum over internal forces by
invoking Newton’s third law. Here the situation is more complicated. To get an
expression that depends only on external forces, we need the double sum over internal
moments in Eq. (7.3) to be zero. We call this the internal-moment assumption.
Assumption 7.1 Under the internal-moment assumption, the sum of all internal
moments among the constituent particles of a multiparticle system is zero:
N N
1 ri/j × Fi,j = 0.
2 i=1 j =1
(7.4)
It is important to remember that not all multiparticle systems satisfy the internalmoment assumption; it is not implied by Newton’s third law. When it is not satisfied,
there is not much more that can be done to model the system other than applying Newton’s second law separately to each particle; examining the total angular momentum
of the collection in that situation may not be very helpful. Fortunately, it is difficult to
248
CHAPTER SEVEN
find a system of particles that does not satisfy this assumption. For almost all the problems and examples in the book (except Problem 7.13, where you can study a simple
counterexample), we assume that the internal-moment assumption is satisfied. In fact,
many systems satisfy the more restrictive condition that the internal force between
every pair of particles in the system is along the line connecting them, so that each
term in the sum in Eq. (7.4) is zero (since the cross product of two parallel vectors is
zero). We call this the internal-force assumption. As always in engineering, it is good
practice to examine your assumptions when attacking a new problem.
When the internal-moment assumption is satisfied, the derivative of the total
angular momentum in Eq. (7.2) simplifies to
Id
dt
I
hO =
N
(ext)
ri/O × Fi(ext) =
MO
,
(7.5)
i=1
(ext)
as the total external moment
where we have used Definition 4.3 to define MO
about O. Eq. (7.5) can be very useful for certain multiparticle problems, particularly
problems in which part of the collection is rotating very fast. It will arise again in the
study of two- and three-dimensional motion of rigid bodies.
Chapter 6 showed that we can treat a collection of particles as a single equivalent
particle of mass mG located at the center of mass. This virtual particle obeys Newton’s
second law, where the total force is the vector sum of all external forces acting on the
collection. The result in Eq. (7.5) is similar, in that it treats the collection as a single
entity and shows that the behavior of its total angular momentum is determined solely
by the total external moment.
BE CAREFUL!
For the translational dynamics of a multiparticle system, the total linear momentum
Ip is equal to the linear momentum Ip
O
G/O of the center of mass relative to O.
However, for the rotational dynamics of such a system, the total angular momentum
Ih in Eq. (7.5) is not the same as the angular momentum Ih
G/O of the center of mass
O
(ext)
relative to O. Nor is MO computed using the total force on the center of mass. That
is why Eq. (7.5) is of less help than it might initially seem, since computing the total
angular momentum and total external moment can be tricky. Also, it only provides
one equation for what may be a multiple-degree-of-freedom problem. Nevertheless,
there are many problems that can be solved with this equation. As shown in the next
section, it is particularly useful when the total external moment is zero. In addition,
the procedure used to derive Eq. (7.5) is quite general and will inform much of our
future analysis.
Example 7.1 A Two-Particle Pendulum
In this example we turn to a variant of the simple pendulum, this time with two particles connected in an equilateral triangle, with the pivot point as shown in Figure 7.2.
The center of mass is along the line connecting the two particles and serves as a convenient reference point. Because the particles are rigidly connected by massless rods,
this system still has only one degree of freedom. We represent that degree of freedom
ANGULAR MOMENTUM AND ENERGY OF A MULTIPARTICLE SYSTEM
249
I
ey
B
O
g
l
r
l
–
2
2
兹苵3 Te + —Te
–—
r
θ
er
l
–
2
1
兹苵3 Fe – —
–—
Feθ
r
ex
l
θ
mP
eθ
1
2
2
mQ
FP/Q
2
FQ/P
–mQgey
–mPgey
(a) Coordinates
(b) Reference frame
(c) Free-body diagrams
Figure 7.2 A two-particle pendulum configured in an equilateral triangle.
by the angle θ of the line from the attachment point to the center of mass relative to
the local vertical. We wish to find the equation of motion for θ.
One approach to this problem is to follow the procedure outlined in Chapter 6:
separate the particles, write the equation of motion for each, including all constraint
and internal forces, and then combine to eliminate the unknown forces and find a
single equation of motion. This approach is challenging and involved. An easier
approach is to consider the total angular momentum about O.
As usual, we introduce an inertial frame I located at O and a polar frame B
with unit vector er directed toward the center of mass, as shown in Figure 7.2b. The
transformation table between these two frames is
er
eθ
ex
ey
sin θ
cos θ
− cos θ
sin θ .
We next write the kinematics of the two particles:
√
3
1
ler − leθ
2
2
√
1
3
I
vP/O =
l θ̇eθ + l θ̇er
2
2
√
1
3
ler + leθ
rQ/O =
2
2
√
1
3
I
vQ/O =
l θ̇eθ − l θ̇er .
2
2
rP/O =
The total angular momentum is found from Definition 7.1,
I
hO = mP rP/O × IvP/O + mQrQ/O × IvQ/O
= (mQ + mP )l 2θ̇e3.
(7.6)
250
CHAPTER SEVEN
We now use Eq. (7.5) to find the equation of motion. First we need to compute the
total external moment acting on the pendulum, which can be done using the transformation array and the external forces in the free-body diagram in Figure 7.2c. Note that
we have shown the tension in each rod (T and F ), but by using a moment approach,
these are irrelevant, as they do not produce a moment about O. We omit the internal
force between the two particles, as this force cancels out in the angular momentum
equation because of the internal-moment assumption. The relevant moments about O
come only from the external gravity forces:
√
1
3
MP/O = rP/O × FP = l cos θmP ge3 −
l sin θmP ge3
2
2
√
3
1
l sin θmQge3.
MQ/O = rQ/O × FQ = − l cos θmQge3 −
2
2
The final equation of motion is now easily found from Eq. (7.5) by differentiating
the total angular momentum in Eq. (7.6) and setting it equal to the sum of the above
two moments:
√
3g
θ̈ +
sin θ = 0.
(7.7)
2 l
Eq.√
(7.7) is also the equation of motion for a simple pendulum with length equal to
2l/ 3 (which is not equal to rG/O ).
It is worth noting that Example 7.1 worked out so well because the pendulum was
actually a rigid body and thus our single angular-momentum equation was adequate
for the single degree of freedom. If we had allowed the masses to move relative to
each other, it would have been more complicated, as more equations would be needed
to fully describe the system. These could have been obtained using the results of
Chapter 6 (by separating the particles). Soon we will explore an alternative approach
to finding the angular momentum of multiparticle systems and discover one of our
most important results.
7.1.2 Conservation of Total Angular Momentum
Section 6.1.2 showed that, in the absence of external forces on a collection of particles,
the total linear momentum is conserved (Law 6.1). This exceptionally important result
is used for solving many problems. For instance, it was fundamental to the analysis of
collisions in Section 6.2. Here we find a parallel result for the total angular momentum
of a collection of particles. Eq. (7.5) shows that, in the absence of external moments
on the collection, the total angular momentum is a constant of the motion.
Law 7.1 Under the internal-moment assumption, the law of conservation of total
angular momentum states that, if the total external moment acting on a system
of particles about the origin O is zero, then the total inertial angular momentum
of the collection about O is a constant of the motion.
It is worth noting that, unlike for conservation of total linear momentum, here we
can still have a non-zero total external force, as long as it produces no moment about
O. A central force is an example of such a situation (see Definition 4.5). It is also
ANGULAR MOMENTUM AND ENERGY OF A MULTIPARTICLE SYSTEM
I
I
251
vQ/O
ey
Q
l
P
O
I
vP/O
ex
Figure 7.3 Two hockey pucks P and Q connected by a massless string of length l.
important to remember that Law 7.1 only applies when the internal-moment assumption holds. Otherwise, a system without any external moments may not conserve total
angular momentum. You have the opportunity to study such a system in Problem 7.13.
Again, always check your assumptions before embarking on a new problem!
Example 7.2 Two Hockey Pucks
As an example of using conservation of total angular momentum, consider the case
of two sliding hockey pucks connected by a massless string, as shown in Figure 7.3.
Using Cartesian coordinates in frame I = (O, ex , ey , ez ), the position and velocity
of each mass are
rP/O = xP ex + yP ey
I
vP/O = ẋP ex + ẏP ey
rQ/O = xQex + yQey
I
vQ/O = ẋQex + ẏQey .
The total angular momentum about O is the sum of the angular momenta of P and
Q about O:
hO = rP/O × mP IvP/O + rQ/O × mQIvQ/O
= mP (xP ẏP − yP ẋP ) + mQ(xQẏQ − yQẋQ) ez .
I
Despite the internal force connecting mass P and Q, the absence of any external
forces leads to conservation of total angular momentum. (Since angular momentum
is a vector, both its magnitude and direction are conserved.) Differentiating IhO . ez
with respect to time and setting the result equal to zero yields
mP (xP ÿP − yP ẍP ) + mQ(xQÿQ − yQẍQ) = 0.
In this case, conservation of total angular momentum provided only one equation—we
would need three more scalar equations to solve for all four unknowns.
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CHAPTER SEVEN
Example 7.3 Sticky Impact with a Pendulum, Revisited
Example 6.4 considered a sticky collision between a moving particle and a stationary
pendulum. We asked for the velocity of the composite particle after the collision when
the particles are stuck together (which is equivalent to asking for the initial swing
rate of the pendulum). That example proved to be a bit involved because the system
was acted on by external forces (gravity and the tension in the pendulum rod), which
prevented a straightforward application of momentum conservation. We revisit it here
using conservation of angular momentum.
We can fruitfully use conservation of angular momentum here because, right up
until the end of the collision, the only forces in the problem act through attachment
point O and thus produce no moment about O. (Of course, after the collision, when
the particles are stuck together, we then have a simple-pendulum problem, which can
be solved using the angular-momentum equation from Section 4.2.1 and the initial
conditions determined here.) Before the collision, the only contribution to the angular
momentum about O is from particle Q:
I
hO (t1) = rQ/O (t1) × IvQ/O (t1)
= −ley × mQ(ẋQ(t1)ex + ẏQ(t1)ey )
= lmQẋQ(t1)ez ,
where l is the length of the pendulum rod. At the instant the particles collide and stick,
the velocity of Q in the ey direction makes no contribution to the angular momentum.
Of course, at the time of impact the pendulum rod applies a significant impulse to
make the ey component of the velocity of Q zero, but it has no effect on the total
angular momentum.
Right after the collision the total angular momentum is given by the same position
at the bottom of the pendulum crossed with the new linear momentum:
hO (t2) = rPQ/O (t2) × (mP + mQ)IvPQ/O (t2)
I
= −ley × (mP + mQ)ẋPQ(t2)ex
= l(mP + mQ)ẋPQ(t2)ez .
Since this must be the same as the angular momentum before the collision, setting
Ih (t ) = Ih (t ) yields the velocity after the sticky impact:
O 1
O 2
ẋPQ(t2) =
mQ
mP + m Q
ẋQ(t1),
which is the same result as Eq. (6.12).
7.2 Angular Momentum Separation
This section revisits the total angular momentum in Definition 7.1 to get an alternative,
and immensely useful, form. Figure 7.1 shows that we can replace the position of each
particle by the vector triad,
ANGULAR MOMENTUM AND ENERGY OF A MULTIPARTICLE SYSTEM
253
ri/O = rG/O + ri/G,
just as in our linear-momentum discussion. We can also replace the velocity of the
ith particle by taking the derivative of this triad to obtain
Id
dt
(ri/O ) = Ivi/O = IvG/O + Ivi/G.
The result is a new equation for the total angular momentum. Using Definition 7.1,
we have
I
hO =
N
mi (rG/O + ri/G) × (IvG/O + Ivi/G)
i=1
=
N
mi rG/O × IvG/O + rG/O ×
N
mi Ivi/G +
i=1 i=1
=0
N
I
= mGrG/O × vG/O +
=IhG/O
i=1
I
mi ri/G × vG/O +
=0
N
mi ri/G × Ivi/O
i=1
N
i=1
mi ri/G × Ivi/G,
=IhG
where we twice set terms to zero using the center-of-mass corollary in Eq. (6.15).
Now we do have the angular momentum IhG/O relative to the origin of an equivalent point mass of mass mG traveling at the velocity of the center of mass. Although
the total angular momentum in Eq. (7.5) does not equal this simple expression, we
have nonetheless been able to pull it out.
The third term in the equation, IhG, is the total angular momentum of all particles
about the center of mass G. We conclude that the total angular momentum of the
system of particles about O separates into the angular momentum of the center of
mass relative to the origin O of the inertial frame plus the total angular momentum
of all particles about their center of mass G. We call this the separation principle.
This result is so important, we rewrite it:
I
hO = IhG/O + IhG,
(7.8)
hG/O =
mGrG/O × IvG/O
(7.9)
where
I
and
I
hG =
N
i=1
mi ri/G × Ivi/G.
(7.10)
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CHAPTER SEVEN
I
ey
G
Q
I
θ
vG/O
P
O
ex
Figure 7.4 Two hockey pucks P and Q connected by a string of length l.
Thus, if we can find the two angular momenta in Eqs. (7.9) and (7.10), it is easy to find
the total angular momentum relative to the origin of an inertial frame. These momenta
are often much easier to calculate or measure than the total angular momentum.
Example 7.4 Two Hockey Pucks, Revisited
As an example of the separation of total angular momentum, consider again the case
of two sliding hockey pucks connected by a massless string, as shown in Figure 7.4.
Using Cartesian coordinates in frame I = (O, ex , ey , ez ), the position and velocity
of the center of mass G are
rG/O =
m P xP + m Q xQ
mP + m Q
ex +
mP yP + mQyQ
=xG
mP + m Q
ey
=yG
and
I
vG/O = ẋGex + ẏGey .
The angular momentum of the center of mass about the origin is
hG/O = rG × mGIvG/O
I
= mG(xGẏG − yGẋG)ez ,
where mG = mP + mQ.
To compute the angular momentum about the center of mass, we use polar frame
B = (G, b1, b2, b3) attached to the center of mass with b1 directed from G to Q at
angle θ relative to ex . Let rQ/G = lQb1, where
lQ = rQ/O − rG/O =
(xQ − xG)2 + (yQ − yG)2
ANGULAR MOMENTUM AND ENERGY OF A MULTIPARTICLE SYSTEM
255
is constant. Similarly, let rP/G = −lP b1. We have
hG = rP/G × mP IvP/G + rQ/G × mQIvQ/G
I
= (−lP b1) × mP (−lP θ̇b2) + (lQb1) × mQ(lQθ̇b2)
2
)θ̇ez ,
= (mP lP2 + mQlQ
where we used b3 = ez .
We are not yet done, however. Next we take the derivative of each term in
Eq. (7.8), as we did to arrive at Eq. (7.5), to find equations of motion for the angular
momenta.
7.2.1 Angular Momentum of the Center of Mass
We start by taking the derivative of the angular momentum of the center of mass
about O:
Id
dt
I
hG/O
=
Id
dt
mGrG/O × IvG/O
Id
= mGIvG/O × IvG/O +mGrG/O ×
dt
I
vG/O .
=0
The first term is zero because the cross product between a vector and itself is zero. For
the second term, we use the result for the center of mass of a collection of particles
in Eq. (6.16) to replace the velocity derivative with the total external force:
Id
dt
I
(ext)
(ext) = MG/O
hG/O = rG/O × FG
.
(7.11)
Eq. (7.11) is a reassuring result and again confirms that we can treat the collection
of particles as an equivalent single particle of mass mG located at the center of mass.
Eq. (7.11) is the angular-momentum form of Newton’s second law applied to the
center of mass of a multiparticle system, just as we found in Chapter 4 for a single
particle. Thus we can solve for the translational motion of the center of mass using
either the force equation in Eq. (6.16) or the moment equation in Eq. (7.11), whichever
is more convenient for the problem at hand.
The motion of the center of mass described by Eq. (7.11) has no dependence on
the motion of the individual particles relative to the center of mass—unless the force
or moment depends upon them, as in Tutorial 6.1. Amazing. What about the other
term in Eq. (7.8)? Now we look at the total angular momentum about the center of
mass.
256
CHAPTER SEVEN
7.2.2 Angular Momentum about the Center of Mass
As done for IhG/O , we now take the derivative of the total angular momentum of the
system of particles about the center of mass:
Id
I
=
hG
dt
=
N
Id
dt
mi ri/G × Ivi/G
i=1
N
mi Ivi/G × Ivi/G +
i=1
N
=0
mi ri/G ×
i=1
Id
I
dt
vi/G .
Now we have to be clever. Recall that Ivi/O = IvG/O + Ivi/G. Thus
N
mi ri/G ×
i=1
Id
I
dt
vi/G =
N
mi ri/G ×
i=1
Id
dt
N
I
vi/O −
i=1
mi ri/G ×
=0
Id
I
dt
vG/O ,
where we again used the center-of-mass corollary in Eq. (6.15). The result is
Id
I
hG =
dt
N
ri/G × mi
i=1
Id
I
vi/O .
dt
Now we can use Newton’s second law on each particle as in Chapter 6. That is, we
replace the inertial acceleration with the forces acting on particle i,
mi
Id
dt
I
vi/O = Fi(ext) +
N
Fi,j ,
j =1
which leaves
Id
dt
I
hG
=
N
ri/G × Fi(ext) +
=
ri/G × Fi,j
i=1 j =1
i=1
N
N
N ri/G × Fi(ext) +
i=1
N N
1 (ri/G − rj/G) × Fi,j .
2 i=1 j =1
=0
We have once again used the internal-moment assumption. The final result is
Id
dt
I
hG =
N
i=1
ri/G × Fi(ext)
=
N
(ext) (ext)
Mi/G
= MG
.
(7.12)
i=1
As with the total angular momentum, as long as the internal-moment assumption
holds, the effect of the internal forces cancel and the dynamics of the angular mo-
ANGULAR MOMENTUM AND ENERGY OF A MULTIPARTICLE SYSTEM
257
mentum of the collection about the center of mass is determined entirely by the total
external moment about G.
Eq. (7.12) describes the motion of the system of particles about their center of mass.
Compare Eq. (7.12) to Eq. (7.11). First we found that we could write the equation of
motion for the center of mass of the system relative to the origin of the inertial frame,
independently of the motion of the particles about their center of mass. We have now
shown, in Eq. (7.12), that we can write the equation of motion for the total angular
momentum of the system of particles relative to their center of mass solely in terms of
the total external moment about the center of mass, independently of the translational
motion of the center of mass. There is complete separation of motion of the center of
mass and motion about the center of mass!
Eqs. (7.11) and (7.12) also imply a corollary to the law of conservation of angular
momentum. If there are no external moments about the center of mass, either angular
momentum of the center of mass about O is conserved or, more importantly, the
angular momentum of the system about the center of mass is, even if the total angular
momentum about O is not conserved. That is, even if there are external forces acting,
resulting in acceleration of the center of mass, and even if they result in a moment
(ext)
= 0. This is so important
about O, IhG is a constant of the motion as long as MG
that we highlight it as a separate law.
Law 7.2 Under the internal-moment assumption, the law of conservation of total
angular momentum about the center of mass states that, if the total external
moment acting on a system of particles about the center of mass G is zero, then
the total inertial angular momentum about G is a constant of the motion.
The separation result is probably one of the most important results in the entire
book and informs all of Parts Three and Four. It allows us to study the rotational
motion of a rigid body (which we model as a collection of many particles with fixed
position relative to the center of mass) without necessarily having to simultaneously
solve for the translational motion of the body’s center of mass. It also helps explain
all sorts of motion, from tops to gyroscopes to satellites.1
Example 7.5 Tethered Satellites
Consider the motion of a pair of identical satellites connected by a massless tether, as
shown in Figure 7.5. Suppose the satellites have mass m and the tether has length 2l.
Each satellite has a thruster that is capable of exerting force F perpendicular to the
tether. If the thrusters fire simultaneously, and for identical durations, the total force
on the center of mass of the satellite pair is zero and the total linear momentum is
conserved. To determine the thruster’s effect on the angular momentum about center
of mass G, we refer to Eq. (7.12). In particular, suppose we would like to determine
the maximum duration that both thrusters can fire without exceeding 2g centripetal
acceleration onboard either satellite (which might be hazardous to the science cargo).
1 One
caveat. This separation is not always entirely complete because a dependence on translation can be
hidden in the external moment in Eq. (7.12). We showed this in Tutorial 6.1 for translation. Another good
example is orbital and attitude motion of a satellite, where a portion of the torque on the satellite can depend
on where it is in earth orbit (particularly for an elliptical orbit). Nevertheless, there is no coupling in the
angular momenta.
258
CHAPTER SEVEN
I
FQ(ext)
ey
Q
l
l
θ
G
P
FP(ext)
O
ex
Figure 7.5 Identical tethered satellites P and Q.
Let B = (G, b1, b2, b3) be a polar frame with origin G and unit vector b1 directed
from G to Q at angle θ relative to ex . The angular momentum about the center of
mass is
hG = rP/G × mIvP/G + rQ/G × mIvQ/G
I
= (−lb1) × m(−l θ̇b2) + (lb1) × m(l θ̇b2) = 2ml 2θ̇b3.
The total external moment about the center of mass is
(ext)
(ext)
= rP/G × FP(ext) + rQ/G × FQ
MG
= (−lb1) × (−F b2) + (lb1) × (F b2) = 2lF b3.
Therefore, the dynamics of the angular momentum about the center of mass are
2ml 2θ̈ = 2lF.
(7.13)
The centripetal acceleration of P is IaP/G = l θ̇ 2b1. Rearranging Eq. (7.13) and
integrating with θ̇(0) = 0 yields
θ̇ =
F
t.
ml
To ensure IaP/G < 2g, the angular rate must satisfy |θ̇| <
t<
√
2g/ l, which implies
m 2gl.
|F |
One interesting application of a system like this is to produce artificial gravity
during long space voyages by connecting a counter mass to a long tether and rotating
the system such that the centripetal acceleration is equal to g.
Example 7.6 Two Hockey Pucks, Take 3
Here we return to the two rotating hockey pucks from Examples 7.2 and 7.4 and
Figure 7.4. Since there are no external forces on the system, the total linear momentum
ANGULAR MOMENTUM AND ENERGY OF A MULTIPARTICLE SYSTEM
259
and total angular momentum are conserved. In addition, the angular momentum of
the center of mass about O is a constant of the motion:
I
hG/O = mG(xGẏG − yGẋG)ez = constant,
where mG = mP + mQ.
Likewise, because the only internal force is along the line connecting the pucks
and there is no external moment, the angular momentum about the center of mass is
conserved by Law 7.2, which implies
I
2
hG = (mP lP2 + mQlQ
)θ̇ez = constant = h0.
(7.14)
Suppose the length of the string, l = lP + lQ, can change. For instance, we could
imagine motors in the hockey pucks or a reel at the center of the string that draws it in
or lets it out. Since the resulting tension in the string is still along the line connecting
the hockey pucks, the internal-moment assumption still holds and the total angular
momentum about G does not change. As a result, the rotation rate θ̇ must change to
maintain constant magnitude of the angular momentum. When the string shortens the
rotation rate speeds up and when the string lengthens it slows down. Every time you
watch ice skaters spin faster as they bring in their arms you are seeing this effect.
In fact, it is rather simple to find an expression for the change in the rate of rotation
˙ Since the center of mass must be fixed
for a constant rate of change in the separation l.
˙ Thus we can write the two
and the string doesn’t stretch, we know that l˙P = l˙Q = l.
distances from the center of mass as
˙ − t0 )
lP (t) = lP (t0) − l(t
˙ − t0).
lQ(t) = lQ(t0) − l(t
The rotation rate as a function of time is thus
θ̇ (t) =
h0
2
.
˙ − t0 ) 2
˙ − t0) + mQ lQ(t0) + l(t
mP lP (t0) + l(t
The rotation rate of the hockey pucks increases (or decreases) as the inverse of time
squared.
7.3 Total Angular Momentum Relative to an Arbitrary Point
Following the development in Section 4.2.2, this section examines the more general
case of computing the total angular momentum of a system of particles relative to
an arbitrary—and perhaps inertially accelerating—point Q = G. Figure 7.6 depicts
the situation. As in Section 4.2.2, we can write the angular momentum of particle i
relative to Q as
hi/Q = ri/Q × mi Ivi/Q.
I
260
CHAPTER SEVEN
B
Q
ri/Q
I
e2
rQ/G
i
ri/G
rQ/O
ri/O
G
j
Fj(ext)
rG/O
O
e1
e3
Figure 7.6 A collection of particles with center of mass G and an arbitrary reference point Q.
Next we sum the individual angular momenta relative to Q to find the total angular
momentum of the collection relative to the point Q:
I
hQ =
N
ri/Q × mi Ivi/Q.
(7.15)
i=1
Eq. (7.15) is similar to the definition of the angular momentum relative to the center of
mass in Eq. (7.10). The question is whether we can find a similarly compact expression
for the dynamics of IhQ as we did for IhG in Eq. (7.12).
The first step is to rewrite the angular momentum about Q in terms of the angular
momentum about G. Replacing ri/Q and Ivi/Q using ri/G = ri/Q + rQ/G (and its
derivative) yields
I
hQ =
N
mi ri/G − rQ/G ×
vi/G − IvQ/G .
I
i=1
Distributing the cross product results in
I
hQ =
N
i=1
N
I
mi ri/G × vi/G −
=IhG
− rQ/G ×
N
i=1
mi ri/G ×IvQ/G
=0
mi Ivi/G +mGrQ/G × IvQ/G,
i=1 =0
where we have twice used the center-of-mass corollary.
ANGULAR MOMENTUM AND ENERGY OF A MULTIPARTICLE SYSTEM
261
Thus our final expression for the angular momentum relative to Q in terms of the
angular momentum relative to the center of mass G is
I
hQ = IhG + mGrQ/G × IvQ/G.
(7.16)
The usefulness of Eq. (7.16) comes from applying Newton’s second law to it.
Recall that taking the derivative of IhG and applying Newton’s second law resulted
in Eq. (7.12); the dynamics of the collection can be considered relative to the center
of mass without regard to the motion of the center of mass. What if, instead, we are
interested in the motion relative to the arbitrary point Q? What are the dynamics of
Ih ? To find out, we take the derivative of Eq. (7.16) to obtain
Q
Id
dt
I
hQ =
Id
dt
I
hG + m G
Id rQ/G × IvQ/G
dt
=0
+ mGrQ/G ×
Id
dt
vQ/O − IvG/O ,
I
where the second term is zero because the cross product of a vector with itself is zero.
Replacing the inertial derivative of the angular momentum about G with the moment
definition from Eq. (7.12), we find
Id
dt
I
hQ =
N
ri/G × Fi(ext) − rQ/G × mGIaG/O + rQ/G × mGIaQ/O .
i=1
We now use the finding from Chapter 6 that the total mass times the acceleration of
the center of mass is equal to the sum of the external forces on the collection:
mGIaG/O =
N
Fi(ext).
i=1
This equation allows us to write the angular-momentum derivative as
Id
dt
I
hQ =
N
ri/G − rQ/G × Fi(ext) + rQ/G × mGIaQ/O .
i=1
Finally, we use the vector triad rQ/G + ri/Q = ri/G again to obtain
N
N
(ext)
ri/G − rQ/G × Fi(ext) =
ri/Q × Fi(ext) =
MQ
,
i=1
i=1
which results in our final expression for the angular-momentum derivative:
Id
dt
(ext)
hQ = M Q
+ rQ/G × mGIaQ/O .
I
(7.17)
262
CHAPTER SEVEN
Compare Eq. (7.17) to Eq. (7.12). The beauty of considering the angular momentum about the center of mass is that its inertial rate of change is equal solely to the
external moment about G. This allows us to consider the orientation of the particle
collection without worrying about the overall translation of the center of mass. If instead we consider the angular momentum relative to some arbitrary point Q, then
Eq. (7.17) shows that the separation of orientation and translation no longer holds.
The rate of change of IhQ depends on the external moment about Q and the inertial
acceleration of Q. Nevertheless, there are problems where this added complication is
offset by the advantage of using a more geometrically convenient point.
Example 7.7 Crane, Take 2
This example revisits the simple model of an overhead crane solved in Example 6.2
and shown in Figure 6.2, but with the addition of a horizontal external force on the
block, FQ = uex . This force can be considered a control force used to move the block
and center the pendulum (this is a common problem in automatic control). We wish to
find the equations of motion for the block and the pendulum. In Chapter 6 we solved
this problem by considering Newton’s second law separately on the block and the
pendulum mass and then eliminating the constraint forces. Here we use the angular
momentum about Q and Eq. (7.17).
Since the block has no angular momentum about Q, the total angular momentum
about Q is equal to the angular momentum of the pendulum mass P about Q,
hQ = mP rP/Q × IvP/Q.
I
Using rP/Q = ler and its inertial derivative IvP/Q = l θ̇eθ gives
I
hQ = mP l 2θ̇e3.
(7.18)
We now insert Eq. (7.18) into Eq. (7.17). The inertial derivative of the angular
momentum about Q is
Id
dt
I
hQ = mP l 2θ̈e3.
The moment about Q is
MQ = rP/Q × FP = ler × (−mP g)ey = −mP gl sin θe3.
To calculate the correction term in Eq. (7.17), rQ/G × mGIaQ/O , we use
˜ r,
rQ/G = rQ/O − rG/O = −le
mP
I
where l˜ =
mG l, as defined in Example 6.6. The acceleration of Q in I is aQ/O = ẍex .
Thus the angular momentum derivative in Eq. (7.17) becomes
mP l 2θ̈e3 = −mP gl sin θe3 − mGl˜ẍer × ex
= −mP gl sin θe3 − mP l ẍ cos θe3.
ANGULAR MOMENTUM AND ENERGY OF A MULTIPARTICLE SYSTEM
263
This equation simplifies to the single scalar equation
l θ̈ + ẍ cos θ + g sin θ = 0.
(7.19)
Eq. (7.19) is only a single equation in the two unknowns ẍ and θ̈. Another equation
comes from examining the center of mass. The results of Chapter 6 indicate that the
acceleration of the center of mass is equal to the total external force acting on the
system, which is uex . (For this problem it is easier to consider the translational motion
of the center of mass using Newton’s second law rather than the angular momentum
form.) Eq. (6.18) of Example 6.6 computed Newton’s second law for the center of
mass of the crane:
Id
dt
pG/O = mG(ẍex + l˜θ̈eθ − l˜θ̇ 2er ) = (N − mGg)ey + uex .
I
The ex component of this equation supplies the second equation of motion,
mGẍ + mGl˜θ̈ cos θ − mGl˜θ̇ 2 sin θ = u.
(7.20)
Eqs. (7.19) and (7.20) are the two equations of motion, arrived at without considering the internal reaction forces. A small amount of algebra leads to the following
separate equations of motion for ẍ and θ̈ of Eqs. (6.5) and (6.6):
mP
m
m
u
2
1−
cos θ ẍ − P g sin 2θ − P l θ̇ 2 sin θ =
mG
2mG
mG
mG
mP
m
g
u
2
1−
cos θ θ̈ + P θ̇ 2 sin 2θ + sin θ = −
cos θ.
2mG
l
mG
mG l
As a check on the validity of these equations, it is interesting to examine what
happens when mQ becomes very large relative to mP (and thus mG mP ). Our
intuition tells us that, as the block becomes very heavy, it will look more and more
like a fixed mass and the pendulum will behave according to the simple pendulum
equations from Chapter 3. For large mQ, mP /mG → 0 and the equations of motion
become approximately
u
mG
u
g
cos θ.
θ̈ + sin θ ≈ −
l
mG l
ẍ ≈
If u = 0, then we obtain the equations of motion of the simple pendulum. If u = 0
and constant, then we have recovered the equation of motion for a simple pendulum
in an accelerating box (see Eq. (4.16) with u = mGa).
7.4 Work and Energy of a Multiparticle System
Section 7.2 showed how the angular momentum of a collection of particles can be
separated into the angular momentum of the center of mass about the origin and the
264
CHAPTER SEVEN
angular momentum of the collection about its center of mass. We also derived separate
equations of motion for each of these two angular momenta. This section performs a
similar analysis to derive work and energy relationships for a system of particles.
7.4.1 Kinetic Energy
Just as we did for the total linear and angular momenta of the collection, we define
the total kinetic energy of a system of particles as the sum of the kinetic energy of
each particle.
Definition 7.2 The total kinetic energy of a system of particles about O is the sum
of the individual kinetic energies:
TO =
N
Ti/O =
i=1
N
N
1
1 I
mi Ivi/O 2.
mi vi/O . Ivi/O =
2 i=1
2 i=1
(7.21)
We use the subscript O to indicate that this is the kinetic energy of the system relative
to O (i.e., it is based on the velocity relative to origin O of the inertial frame).
Substituting for Ivi/O using the triad
I
vi/O = IvG/O + Ivi/G
yields
TO =
N
N
N
1 I
1 I
mi vi/G . Ivi/G.
mi Ivi/G +
mi vG/O . IvG/O + IvG/O .
2
2 i=1
i=1
i=1 =0
The second term is zero according to (the derivative of ) the center-of-mass corollary
in Eq. (6.15). The result is another expression for the total kinetic energy,
TO =
N
N
1
1
mi IvG/O 2 +
mi Ivi/G2 .
2 i=1
2 i=1
=TG/O
(7.22)
=TG
Just as for the total angular momentum, the total kinetic energy separates into a
kinetic energy TG/O of the center of mass and a kinetic energy TG about the center of
mass:
TO = TG/O + TG,
(7.23)
1
mGIvG/O 2
TG/O =
2
(7.24)
where
ANGULAR MOMENTUM AND ENERGY OF A MULTIPARTICLE SYSTEM
265
and
N
1
TG =
mi Ivi/G2.
2 i=1
(7.25)
Note that this notation might be a bit confusing, since, as we have discussed, the
velocity is not measured relative to a point but to a frame. We are not measuring the
velocity relative to a frame fixed to G; on the contrary, all of the velocities are inertial,
as required by Definition 5.3 of kinetic energy in Chapter 5. TG is simply that part of
the total kinetic energy due to the change in time of the position of each point relative
to the center of mass of the particle collection.
Example 7.8 Separation of Kinetic Energy of Two Hockey Pucks
This example computes the total kinetic energy of two sliding hockey pucks connected
by a massless string, as shown in Figure 7.4. Recall from Example 7.4 that the position
and velocity of the center of mass G are
rG/O =
m P xP + m Q xQ
mP + m Q
ex +
m P yP + m Q yQ
mP + m Q
=xG
ey
=yG
and
I
vG/O = ẋGex + ẏGey .
Therefore the kinetic energy of the center of mass with respect to O is
1
1
2
+ ẏG)2.
TG/O = mGIvG/O 2 = mG(ẋG
2
2
To compute the kinetic energy about the center of mass, we use rP/G = −lP b1 and
rQ/G = lQb1, where b1 is the unit vector of the polar frame B = (G, b1, b2, b3) that
is directed from G to Q at angle θ with respect to ex (see Figure 7.4). We have
1
1
1
2
)θ̇ 2.
TG = mP lP θ̇b22 + mQlQθ̇b22 = (mP lP2 + mQlQ
2
2
2
The total kinetic energy of the two-particle system is, of course, TO = TG/O + TG.
7.4.2 The Work–Kinetic-Energy Formula
What about the work–kinetic-energy relationship for a system of particles? Recall the
work–kinetic-energy formula from Chapter 5 (Eq. (5.6)) that relates the total work
Wi(tot) done on particle i to its change in kinetic energy:
Ti/O .
Wi(tot)(ri/O (t); γi ) = Ti/O (t2) − Ti/O (t1) =
266
CHAPTER SEVEN
We introduce the notation to concisely indicate the change in kinetic energy from
time t1 to t2. The total work W (tot) of a multiparticle system is defined as the sum of
the work on each particle:
W (tot)({ri/O ; γi }N
i=1) =
N
Wi(tot)(ri/O (t); γi )
i=1
=
N
Ti/O = TO .
(7.26)
i=1
Note that we used the definition of the total kinetic energy. Also, the notation
{ri/O ; γi }N
i=1 is short for {r1/O , . . . , rN/O ; γ1, . . . , γN }.
Let us examine the total-work term more carefully. Using Definition 5.1 for work
from Chapter 5, we can write the total work as the sum of the work done by internal
and external forces, which we call the internal and external work:
W
(tot)
({ri/O ; γi }N
i=1) =
N i=1
γi
Fi(ext) . Idri/O
(ext)
=Wi
+
N
N i=1
(ri/O ;γi )
γi j =1
Fi,j . Idri/O .
(int)
=Wi
(ri/O ;γi )
The total work is just the sum of the internal and external work on each particle over its
trajectory γi . Sometimes it is convenient to make the same substitution as we did for
the angular-momentum calculation (in going from Eq. (7.2) to Eq. (7.3)) and rewrite
the second term in an alternate form:
W (tot)({ri/O ; γi }N
i=1) =
N i=1
γi
Fi(ext) . Idri/O +
=W (ext) ({r
N
i/O ;γi }i=1)
N N 1 Fi,j . Idri/j .
2 i=1 j =1 γi
(7.27)
=W (int) ({ri/O ;γi }N
)
i=1
This form also illustrates an important fact about the work in multiparticle systems.
When we calculated the total moment for the angular-momentum equation, the second
term involving the double sum of internal forces was identically zero because of the
internal-moment assumption. Here it is not zero because we have a dot product rather
than a cross product. The second term in Eq. (7.27) represents the work done against
internal forces when the particles move relative to one another. The existence of
this term should make intuitive sense. Imagine a rubber sheet being deformed and
stretched. Clearly, work is being done against the internal tension forces to stretch the
sheet, that is, to move the particles apart. A common mistake is to forget the internal
work in a multiparticle system!
We write the work–kinetic-energy formula for a system of particles as
W (tot) = TO (t2) − TO (t1) = W (ext) + W (int).
(7.28)
ANGULAR MOMENTUM AND ENERGY OF A MULTIPARTICLE SYSTEM
B
I
ey
Tey
Nb2
b1
g
267
Tb1
b2
mA
s
–μcNb1
mB y
mA
mB
–mAgey
–mBgey
θ
O
ex
(a) Coordinates and reference frames
(b) Free-body diagrams
Figure 7.7 Pulley on a ramp.
To simplify the notation, we have dropped the arguments. The external and internal
work are, respectively,
W (ext)({ri/O ; γi }N
i=1) =
N i=1
γi
Fi(ext) . Idri/O
and
W
(int)
({ri/O ; γi }N
i=1) =
N N 1 Fi,j . Idri/j .
2 i=1 j =1 γi
Eq. (7.28) is the work–kinetic-energy formula for a multiparticle system. We use it
the same way we used the single-particle version: to make predictions about the state
of the system at two different times. Note that we can also separate the total work
into conservative and nonconservative parts. For example, the total work performed
by conservative forces from time t1 to time t2 is W (c)({ri/O (t1), ri/O (t2)}N
i=1), or
W (c)(t1, t2) for short.
Example 7.9 Pulley on a Ramp
Consider two blocks A and B connected by a rope of fixed length, as shown in
Figure 7.7a. We assume that the blocks are initially at rest.
Let μc denote the coefficient of friction between block A and the ramp. Suppose
the blocks are released from rest at time t1 and mB > mA. In this example we use workenergy methods to find their speed at time t2. The reference frames I = (O, ex , ey , ez )
and B = (O, b1, b2, b3) are related by the transformation table
ex
b1
b2
cos θ
− sin θ
ey
sin θ
cos θ .
268
CHAPTER SEVEN
Using the coordinates shown in Figure 7.7a, the kinematics of block A are
rA/O = sb1
I
vA/O = ṡb1
I
aA/O = s̈b1
and, using the kinematic constraint ṡ = −ẏ imposed by the rope, we have
rB/O = x0ex + yey
I
vB/O = −ṡey
I
aB/O = −s̈ey .
The total force on block A, from the free-body diagram in Figure 7.7b, is
FA = (T − μc N )b1 + N b2 − mAgey ,
and the total force on block B is
FB = (T − mB g)ey .
Applying Newton’s second law to block A yields
(T − μc N )b1 + N b2 − mAgey = mAs̈b1.
Using the transformation table and equating forces in the b2 direction gives the normal
force:
N = mAg cos θ.
Using IdrA/O = IvA/O dt and IdrB/O = IvB/O dt, the total work on the system of
two blocks during the period from t1 to time t2 is
W
(tot)
FA . IdrA/O +
(rA/O , rB/O ; γA, γB ) =
=
FB . IdrB/O
γB
γA
s(t2 )
((T − μc mAg cos θ)b1 + mAg cos θb2 − mAgey )
s(t1)
. (dsb1) +
s(t2 )
((T − mB g)ey ) . (−dsey )
s(t1)
= (−μc mAg cos θ − mAg sin θ + mB g)s ,
=W (ext) (rA/O ,rB/O ;γA ,γB )
ANGULAR MOMENTUM AND ENERGY OF A MULTIPARTICLE SYSTEM
N1b2
B
I
b2
mA
s
O
N2b2
Tb1
mB
ey
269
mA
g
b1
–μcN2b1
θ
(a) Coordinates and reference frames
mB
μcN2b1
–N2b2
–mAgey
ex
Tb1
–mBgey
(b) Free-body diagrams
Figure 7.8 Pulley on a ramp, revisited.
where s = s(t2) − s(t1). Note that, in this example, the total internal work done on
the system is zero. By Eq. (7.28), the change in total work on the system equals the
change in the total kinetic energy:
1
1
1
W (tot)(rA/O , rB/O ; γA, γB ) = mAṡ(t2)2 + mB ṡ(t2)2 = (mA + mB )ṡ 2(t2),
2
2
2
where we used the fact that the kinetic energy at t1 is zero. If we also assume that
s(t1) = 0, then
2(−μc mAg cos θ − mAg sin θ + mB g)s(t2)
.
ṡ(t2) =
mA + m B
Example 7.10 Pulley on a Ramp, Take 2
Now consider the system of two blocks A and B (masses mA and mB , respectively)
connected by a rope of fixed length L, as shown in Figure 7.8a. Let μc denote the
coefficient of friction between block A and block B, and assume there is no friction
between A and the ramp. Suppose that the blocks are released from rest at time t1 and
mA > mB . In this example we use work-energy methods to find their speed at time
t2. Unlike in the previous example, the total internal work of the system is not zero.
The reference frames I = (O, ex , ey , ez ) and B = (O, b1, b2, b3) are related by the
same transformation table as in the previous example. Using the coordinates shown
in Figure 7.8a, the kinematics of block A are
rA/O = sb1
I
vA/O = ṡb1
I
aA/O = s̈b1,
and, assuming the pulley is located at Lb1, we have
rB/O = (L − s)b1
I
vB/O = −ṡb1
I
aB/O = −s̈b1.
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CHAPTER SEVEN
From the free-body diagram in Figure 7.8b, the total force on block A is2
FA = (T − μc N2)b1 + (N1 − N2)b2 − mAgey ,
and the total force on block B is
FB = (T + μc N2)b1 + N2b2 − mB gey .
Applying Newton’s second law to block B yields
(T + μc N2)b1 + N2b2 − mB gey = −mB s̈b1.
Using the transformation table and equating forces in the b2 direction gives the internal force between the blocks:
N2 = mB g cos θ.
Using IdrA/O = IvA/O dt and IdrB/O = IvB/O dt, the total work on the system of
two blocks during the period from t1 to t2 is
FA . IdrA/O +
FB . IdrB/O
W (tot)(rA/O , rB/O ; γA, γB ) =
γA
s(t2 )
=
γB
((T − μc mB g cos θ)b1 + (N1 − N2)b2 − mAgey ) . (dsb1)
s(t1)
+
s(t2 )
((T + μc mB g cos θ)b1 + N2b2 − mB gey ) . (−dsb1)
s(t1)
= −2μc mB g cos θs + (mB − mA)g sin θs ,
=W (int) (rA/O ,rB/O ;γA ,γB )
=W (ext) (rA/O ,rB/O ;γA ,γB )
where s = s(t2) − s(t1). Note that, in this example, the total internal work on the
system is not zero. By the work-energy formula in Eq. (7.28), the change in total work
on the system equals the change in the total kinetic energy:
1
1
1
W (tot)(rA/O , rB/O ; γA, γB ) = mAṡ(t2)2 + mB ṡ(t2)2 = (mA + mB )ṡ 2(t2).
2
2
2
Assuming s(t1) = 0, we find that
(2(mB − mA)g sin θ − 4μc mB g cos θ)s(t2)
ṡ(t2) =
.
mA + m B
Note that, if we had ignored the internal work in this example, we would have gotten
the wrong answer!
We conclude this section by looking at the power expended by a collection of
forces. Recall from Definition 5.2 that power is defined as the rate of change of work.
2 The internal friction forces are oriented as though s increases; although we don’t know this a priori, you
get the same answer either way.
ANGULAR MOMENTUM AND ENERGY OF A MULTIPARTICLE SYSTEM
271
We showed that power is also equal to the rate of change of kinetic energy. Thus
using the definition of total work in Eq. (7.26) and the work–kinetic-energy formula
in Eq. (7.28), the total power acting on the collection of particles can be written as
P(tot) =
N
Fi(ext) . Ivi/O +
i=1
N N
1 d
Fi,j . Ivi/j = TO .
2 i=1 j =1
dt
If we simply replace Ivi/O with IvG/O + Ivi/G, we can write the power due to the
external forces as a sum of the power due to the motion of the center of mass plus the
power due to motion relative to the center of mass:
(ext) . I
P(tot) = FG
vG/O +
N
Fi(ext) . Ivi/G +
i=1
N N
1 Fi,j . Ivi/j .
2 i=1 j =1
(7.29)
The first term in Eq. (7.29) looks like the power associated with a single equivalent
particle of mass mG located at the center of mass of the collection. This leads us
to ask whether the center of mass satisfies the same energy conservation laws as a
single particle, just as it satisfies Newton’s second law (Section 6.1) and the angularmomentum form (Section 7.3). Recall from Chapter 5 that we found the relationship
between power and the change in kinetic energy by differentiating the kinetic-energy
formula. We do that again here, using the separation property of kinetic energy and
differentiating the kinetic energy of the center of mass in Eq. (7.24):
1
d
(TG/O ) = mG
dt
2
Id dt
Id I
vG/O . IvG/O + IvG/O .
vG/O
dt
I
Id I
= mG
vG/O .IvG/O .
dt (ext)
=FG
Chapter 6 showed that we can apply Newton’s second law to the center of mass, so
I (ext)
we replace mG dtd IvG/O with the net force FG
acting on the center of mass G to
find
d
(ext) . I
TG/O (t) = FG
vG/O (t) = PG/O ,
dt
(7.30)
where PG/O is the power associated with total force acting on the center of mass.
This is a satisfying result. Eq. (7.30) shows that the first term of Eq. (7.29) is indeed
the power associated with motion of just the center of mass. It also shows that, just
as for linear and angular momentum, we can treat the kinetic energy of the center of
mass of the collection as if it were the kinetic energy of a single equivalent particle,
and it satisfies the same power relationship. In particular, if the total external force is
zero, the kinetic energy of the center of mass is a constant (like the linear momentum),
even if there are internal or external forces acting on the individual particles and even
if each particle is accelerating. This result will prove very useful in the study of rigid
bodies.
272
CHAPTER SEVEN
7.4.3 Total Energy
Just as in Chapter 5, when we discussed the energy of a single particle, we can now use
the potential energy when either the external or internal work involves conservative
(Fi,j )
(ext)
N
forces. Thus we let Ui/O
(ri/O ) and Ui(int)(ri/1, . . . , ri/N ) =
j =1 Ui/j (ri/j ) denote the scalar potentials corresponding to external and internal conservative forces,
respectively, on particle i. Recalling the definition of potential energy, we have
(ext)
Fi(c,ext) = −∇Ui/O
(ri/O )
(F )
(c)
Fi,j
= −∇Ui/j i,j (ri/j ).
Using Eq. (7.27), the total conservative work is
W
(c)
({ri/O ; γi }N
i=1) =
N i=1
γi
N N 1 (c) . I
dri/j . (7.31)
+
Fi,j
2 i=1 j =1 γi
Fi(c,ext) . Idri/O
)
=W (c,ext) ({ri/O ;γi }N
i=1
)
=W (c,int) ({ri/O ;γi }N
i=1
Now we rewrite the total conservative work in Eq. (7.31) using the change in total
potential energy for the conservative forces:
W (c)(t1, t2) = UO(ext)(t1) − UO(ext)(t2) + UO(int)(t1) − UO(int)(t2),
(ext)
=−UO
(7.32)
(int)
=−UO
where
UO(ext)(t) =
N
(ext)
Ui/O
(ri/O (t))
(7.33)
i=1
UO(int)(t) =
N N
1 (Fi,j )
U
(ri/j (t)).
2 i=1 j =1 i/j
(7.34)
Eq. (7.32) leads to our final work-energy formula for a multiparticle system. We
define the total energy as the sum of the potential and kinetic energies, only now of
the system of particles:
EO (t) =
TO (t) + UO(ext)(t) + UO(int)(t) = TG/O (t) + TG(t) + UO(ext)(t) + UO(int)(t).
The work-energy formula then looks just like Eq. (5.16):
EO (t2) = EO (t1) + W (nc,ext) + W (nc,int).
(7.35)
Thus Eq. (7.35) also necessitates an updated law of conservation of total energy.
ANGULAR MOMENTUM AND ENERGY OF A MULTIPARTICLE SYSTEM
273
Law 7.3 The law of conservation of total energy of a multiparticle system states
that, if the total nonconservative force is zero, the total energy of a multiparticle
system is a constant of the motion.
This law also suggests a power form, as in Chapter 5. Using the same line of
reasoning, the power associated with the nonconservative forces is equal to the rate
of change of the total energy:
N N
N
d
1 (nc) . I
vi/j .
Fi,j
EO = P(nc) =
Fi(nc,ext) . Ivi/O +
2
dt
i=1 j =1
i=1
Example 7.11 Dissipation of Total Energy of a Pair of Sprung Masses
Consider a pair of identical particles of mass m on a horizontal surface connected by
(a) a spring with constant k and rest length r0 and (b) a viscous damper, as shown
in Figure 7.9. There are no external (horizontal) forces acting on either mass, which
implies that TG/O and UO(ext) are constant. There is, however, nonconservative internal
work due to the damper, which will eventually bring the distance between the masses
to r0, regardless of their initial conditions. The amount of work done by the damper is
W (nc,int) = EO (∞) − EO (0) = TG(∞) − TG(0) + UO(int)(∞) − UO(int)(0).
In this example we compute W (nc,int).
Consider a polar frame B = (G, er , eθ , ez ) attached to the center of mass G, with
unit vector er directed from G to Q at angle θ from e1. The distance between the
masses is r. We have
r
rP/G = − b1
2
ṙ
r
I
vP/G = − b1 − θ̇b2
2
2
and similarly for mass Q for which rQ/G = 2r b1. The total internal potential energy is
1
UO(int) = k(r − r0)2,
2
I
e2
Q
r
θ
P
O
e1
Figure 7.9 A pair of sprung masses.
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CHAPTER SEVEN
and the kinetic energy about the center of mass is
1
1
m 2
TG = mIvP/G2 + mIvQ/G2 =
ṙ + r 2θ̇ 2 .
2
2
4
The latter expression can be simplified using conservation of angular momentum
about the center of mass:
I
hG = rP/G × mIvP/G + rQ/G × mIvQ/G =
mr 2θ̇ ez = mhGez ,
2
where hG = r(0)2θ̇(0)/2 is constant.
Using r(∞) = r0 and ṙ(∞) = 0, we have θ̇(∞) = hG/r02 and
W (nc,int) =
mh2G
4r02
−
m
4
ṙ(0)2 +
h2G
r(0)2
1
− k(r(0) − r0),
2
which depends only on θ̇(0), r(0), and ṙ(0).
7.5 Tutorials
Tutorial 7.1 The Two-Body Problem
Chapter 4 developed the equation of motion of a body acted on by a central gravitational force, where we assumed that the central attracting body was fixed in the inertial
frame. In this chapter, we relax that assumption and treat the more realistic problem
of two gravitating bodies (i.e., two very big particles) free to move in absolute space.
This problem is depicted in Figure 7.10. For illustration, we have chosen m2 m1,
such as the sun and a planet or the earth and a satellite. Although the mass disparity
is a common scenario, the result presented here is exact for two attracting bodies of
any mass.
The key assumption is that there are no external forces acting on the bodies;
the only force is the internal gravitational force acting along the line between them
(parallel to the vector r1/2 in Figure 7.10). Therefore the center of mass G is either
fixed or moving at a constant velocity in the inertial frame; the acceleration IaG/O
is zero. In addition, the total angular momentum IhO from Definition 7.1 is constant
because its inertial derivative in Eq. (7.5) is zero. In fact, the same is true for both
the angular momentum IhG/O of the center of mass G about the origin O and the
angular momentum IhG of the two bodies about the center of mass G, whose inertial
derivatives are given in (7.11) and (7.12), respectively. (The latter claim is true
because the internal gravitational force is parallel to r1/2.) Because the total angular
momentum IhO is fixed, the bodies must move in a planar orbit that is orthogonal to
Ih , just as in Example 4.9.
O
These observations indicate that the two bodies orbit about G such that the total
angular momentum relative to G is constant. This result is slightly different from the
picture given in Example 4.9, where the small body orbited the attracting origin. It is
ANGULAR MOMENTUM AND ENERGY OF A MULTIPARTICLE SYSTEM
275
m2
I
e2
G
eθ
r2/O
r1/2
rG/O
G θ
r1/O
O
m1
e1
e3
B
r
(a)
er
(b)
Figure 7.10 Two bodies of masses m1 and m2 in orbit about their common center of
mass in the plane orthogonal to e3. (a) Inertial frame I = (O, e1, e2 , e3). (b) Polar frame
B = (G, er , eθ , e3).
also perhaps different from our intuitive image of the earth orbiting the sun. In fact,
the earth and sun orbit their common center of mass; because the sun is so much
larger than the earth, the center of mass of the two bodies is actually inside the sun.
Consequently, the sun’s motion is very small.3 (The same is true of the earth’s motion
when it is orbited by a small satellite.) Nevertheless, it is very common and useful
to find a description of the earth’s motion relative to the sun (or, to be more general,
body 1 relative to body 2) given by r1/2(t).
Using the inertial frame I = (O, e1, e2, e3), we write the dynamics of body 1
relative to body 2. We start with the definition of the center of mass,
rG/O =
m2
m1
r1/O +
r2/O ,
m1 + m 2
m1 + m 2
=μ1
=μ2
where we have introduced the dimensionless masses μ1 =
m1/(m1 + m2) and μ2 =
m2/(m1 + m2). Note μ1 + μ2 = 1 and m1 + m2 = mG. Rearranging and using the
vector triad r2/O = r2/1 + r1/O , we find
μ1r1/O = rG/O − μ2r2/O = rG/O − μ2(r2/1 + r1/O )
= rG/O + μ2r1/2 − μ2r1/O ,
which means
r1/O = rG/O + μ2r1/2.
3 Scientists searching for a planet about another star infer the presence of a planet by measuring the (small)
motion of the star.
276
CHAPTER SEVEN
Consequently, the kinematics of body 1 are
v1/O = IvG/O + μ2Iv1/2
I
a1/O = IaG/O +μ2Ia1/2,
I
=0
where the absence of external forces implies that the acceleration of G is zero. Using
Newton’s second law, we find
F1,2 = m1Ia1/O = m1μ2Ia1/2,
(7.36)
where F1,2 is the gravitational force on body 1due to body 2. Using Newton’s universal
law of gravity for F1,2 (see Eq. (4.25)) we obtain the two-body vector equation of
motion,
m1m2 I
−Gm1m2
a1/2 =
r1/2,
m1 + m 2
r1/23
or more simply,
Id 2 dt 2
GmG
r1/2 +
r1/2 = 0.
r1/23
(7.37)
Eq. (7.37) is the same equation of motion as in Eq. (4.26) except with mO replaced
by mG = m1 + m2 and P/O replaced by 1/2. The solution is thus the same (elliptical
orbits of m1 about m2) but with slightly different periods.
As in Example 4.9, we can also examine the two-body problem in polar coordinates
using components in polar frame B = (G, er , eθ , e3), which is fixed to G and rotating
so that er is directed toward body 1. Let r =
r1/2 denote the distance between body
1 and body 2, and let θ denote the angular separation between frame B and frame I .
(Assume θ increases in the direction of rotation of the bodies.) We have
r1/2 = rer
I
v1/2 = ṙer + r θ̇ eθ
I
a1/2 = (r̈ − r θ̇ 2)er + (2ṙ θ̇ + r θ̈)eθ .
The gravitational force F1,2 is
F1,2 = −
Gm1m2
er .
r2
Substituting the expressions for F1,2 and Ia1/2 into Eq. (7.36) yields
2ṙ θ̇
r
Gm
r̈ = − 2 G + r θ̇ 2,
r
θ̈ = −
(7.38)
(7.39)
ANGULAR MOMENTUM AND ENERGY OF A MULTIPARTICLE SYSTEM
277
which are the same equations of motion as in Example 4.9, again with mO replaced
by mG.
Eqs. (7.38) and (7.39) are not the only results we have seen before. For example,
computing the angular momentum IhG of bodies 1 and 2 about G gives
I
hG =
2
mi ri/G × Ivi/G,
i=1
where
r1/G = μ2rer
I
v1/G = μ2(ṙer + r θ̇eθ )
r2/G = −μ1rer
I
v2/G = −μ1(ṙer + r θ̇eθ ).
Consequently, we have
I
hG =
2m1m2 2
r θ̇e3,
m1 + m 2
which implies that the magnitude of the angular momentum is proportional to r 2θ̇—
the same value as the specific angular momentum hP in Tutorial 4.2!
In fact, all of the results from Example 4.9 and Tutorial 4.2 apply here, as well as
those from Tutorial 5.2. The orbit of body 1 about body 2 is still a conic section. We
can study orbits about the sun with no error or approximations, as long as we replace
mO in Example 4.9 with mG, the sum of the two masses. This tutorial also shows
why the analysis in Example 4.9, where we assumed the central body was fixed in
absolute space, was a good approximation: if m1 m2, we introduce little error by
using m1 instead of mG.
Tutorial 7.2 The Energy of Two Falling Masses Connected by a Spring
This tutorial revisits the problem examined in Tutorial 6.1 of two falling masses
connected by a spring. We showed that, even though the dynamics of the center
of mass separates from the motion about the center of mass, the full equations of
motion do not always separate. In Tutorial 6.1 we could treat the center-of-mass
motion completely independently of the motion relative to the center of mass as long
as the gravitational field was a constant; this was not the case when we treated gravity
more precisely using the 1/r 2 form. Here we look at the total energy of this nonrigid
collection of particles.
First we write the kinetic energy as the sum of the individual kinetic energies,
1 2
1
2
TO = mẏm
+ M ẏM
2
2
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CHAPTER SEVEN
or equivalently, as the sum of the kinetic energy of the center of mass plus the kinetic
energy due to motion relative to the center of mass,
1
1 2
1
2
2
+ mẏm/G
+ M ẏM/G
.
TO = (m + M)ẏG
2
2
2
What about the work done on the masses? The work done by the internal force
(the spring) is equal to the change in the potential energy of the spring between the
endpoints,
2 1 2
1 W (int) = k l − (ym(tf ) − yM (tf )) − k l − (ym(t0) − yM (t0)) ,
2
2
which can also be written in terms of the motion relative to the center of mass,
2 1 2
1 W (int) = k l − (ym/G(tf ) − yM/G(tf )) − k l − (ym/G(t0) − yM/G(t0)) .
2
2
The external work is the work done in moving the masses down in the gravitational
field. For the constant-gravity model, the work on each mass is
(ext)
= mg(ym(tf ) − ym(t0))
Wm/O
(ext)
= Mg(yM (tf ) − yM (t0)).
WM/O
Adding these work sources together gives the total work,
(ext)
(ext)
+ WM/O
+ W (int)
W (tot) = Wm/O
2
1 = (M + m)g(yG(tf ) − yG(t0)) + k l − (ym/G(tf ) − yM/G(tf ))
2
=WG/O
2
1 − k l − (ym/G(t0) − yM/G(t0)) .
2
For the constant-gravity case, the total work is the work done on the center of
mass by gravity (WG/O ) plus the internal work of the spring due to motion relative to
the center of mass. In fact, because the force of gravity is a constant, we could have
computed the work done on the center of mass directly. Since all of the forces, whether
internal or external, are conservative, we can also write the following conservation
law for the center of mass:
1
2
= constant.
EG/O = (M + m)gyG + (M + m)ẏG
2
ANGULAR MOMENTUM AND ENERGY OF A MULTIPARTICLE SYSTEM
279
The situation is different for the 1/r 2 gravity field. While the internal work is the
same, the external work due to gravity, which equals the change in potential energy,
is now equal to (see Table 5.1)
W (ext) = μm
1
1
−
RE + ym(tf ) RE + ym(t0)
+ μM
1
1
−
,
RE + yM (tf ) RE + yM (t0)
where μ = GmE . This work expression cannot be separated into the work associated
with the position of the center of mass and that associated with the motion relative to
the center of mass (even though we can perform this separation for the kinetic energy).
Consequently, we cannot apply a conservation-of-energy law solely to the center-ofmass motion. This failure to separate is due to the specific nonlinear nature of the
gravitational force.
Nevertheless, we can obtain an expression for the conservation of total energy,
since all the forces are conservative. The total energy of the system is
μm
μM
1 2
1
2
+
+
+ M ẏM
EO = mẏm
Re + y m R e + y M
2
2
2
1 + k l − (ym − yM ) = constant.
2
Tutorial 7.3 Scattering
Chapter 6 examined in detail the collision problem of two solid particles colliding and rebounding with altered trajectories. In this tutorial we discuss a similar
problem called scattering. Particles interact through field forces, such as gravity or
electrostatics, and thus change their trajectories as they come into proximity of each
other without actually colliding. We have already seen one example of scattering—
hyperbolic orbits in the two-body problem. In that case a small body approaches from
an infinite distance, interacts gravitationally with a large body, and moves off to infinity at some angle relative to the arrival trajectory. We derive the equations for that
trajectory in more detail here.
Scattering is an important problem in physics and describes common dynamical
behavior from the very small atomic scale to the large-scale motion of comets and
other planetary bodies. In fact, the famous Rutherford experiment verified the atomic
nature of matter by scattering charged particles off the nucleus.4
In this tutorial we consider two particles, P and Q, moving toward each other
from an infinite separation on straight-line trajectories, each at the same constant
speed v∞ in an inertial frame I , as shown in Figure 7.11a. Their trajectories are
separated by a distance b, called the impact parameter. They interact through a
4 Ernest Rutherford (1871–1937) was a New Zealand chemist and physicist. He is credited as the discoverer
of the nucleus from his famous gold-foil experiment, where he measured the scattering of a beam of alpha
particles directed at a thin gold foil. He was awarded the Nobel Prize in chemistry in 1908.
280
CHAPTER SEVEN
e2
I
I
vP/O(t0) = v ∞e1
e1
P
O
FP,Q
C
b
θP
P
G
I
vQ/O(t 0) = –v ∞e1
FQ,P
Q
(a) Reference frames and geometry
Q
(b) Free-body diagram
Figure 7.11 (a) Two charged particles traveling along straight-line paths toward each other
with separation b. (b) The free-body diagram of each particle.
repulsive electrostatic force along the line between them. We are interested in the
trajectory of each particle relative to the inertial frame C located at their center of
mass. Since there are no external forces acting, the center of mass is either fixed or
moving at constant velocity. The center of mass is located on the line connecting the
particles, as shown in Figure 7.11a, and its position in I is given by
rG/O =
1
mP rP/O + mQrQ/O .
mP + m Q
The (constant) velocity of the center of mass in I can be found using the initial velocity
of the two particles (their velocities at infinity):
I
vG/O (t) =
1
mP IvP/O (t0) + mQIvQ/O (t0)
mP + m Q
=−
mQ − mP
mP + m Q
v∞e1.
(7.40)
Let mQ ≥ mP , which implies the center of mass moves in the −e1 direction unless
mQ = mP , in which case G is fixed.
We use Newton’s second law to find separately the equations of motion for the
position of each particle relative to the center-of-mass frame C. The repulsive force
between the particles in the free-body diagram in Figure 7.11b is given by Coulomb’s
law,
FP ,Q = −
KrQ/P
rQ/P 3
= −FQ,P ,
ANGULAR MOMENTUM AND ENERGY OF A MULTIPARTICLE SYSTEM
281
where K > 0 is a constant proportional to the charges on the particles. The position
of Q relative to P is
rQ/P = rQ/G − rP/G.
The equations of motion for each particle in C are
Cd 2
KrQ/P
mP 2 rP/G = −
dt
rQ/P 3
mQ
Cd 2
dt
r
=
2 Q/G
KrQ/P
rQ/P 3
.
We now use the fact that no external forces are acting on the system. The center
of mass is thus moving at the constant velocity given in Eq. (7.40) and the relative
positions of the two particles satisfy the center-of-mass corollary, which implies
rQ/G = −
mP
rP/G.
mQ
This equation allows us to rewrite the separation between the particles as
rQ/P = rQ/G − rP/G = −
mP + mQ
mP
rP/G − rP/G = −
mQ
mQ
rP/G.
Performing a similar calculation for rQ/G yields the following equations of motion
in C:
Cd 2
μP rP/G
dt
rP/G3
r
−
2 P/G
Cd 2
rQ/G
dt 2
−
μQrQ/G
rQ/G3
=0
(7.41)
= 0,
(7.42)
where
mQ
K
μP =
mP
mP + m Q
K
μQ =
mQ
mP
mP + m Q
2
2
.
The absence of an external force allowed us to rewrite the equations of motion as
two independent equations for each of the particles P and Q relative to the center of
mass. Because of the symmetry of Eqs. (7.41) and (7.42), solving one generates the
solution to the other. Let us solve for the motion of P relative to the center of mass.
Eq. (7.41) has the same form as the orbit equation in Eq. (4.26), which means we can
follow a similar procedure for solving it. Consider the position of P with respect to
G in polar coordinates (rP , θP )C . This choice of coordinates allows us to separate the
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CHAPTER SEVEN
equation of motion into two scalar differential equations:
θ̈P = −
2ṙP θ̇P
rP
(7.43)
μP
+ rP θ̇P2 .
rP2
r̈P =
(7.44)
We now use the fact that the force between the particles acts along the line
connecting them. Thus the internal-moment assumption holds and the total angular
momentum about G is conserved:
hG = mP rP/G × CvP/G + mQrQ/G × CvQ/G = constant.
C
(7.45)
We can calculate the total angular momentum using the initial conditions when the
particles are infinitely far apart. Since in I , the particles have equal and opposite
velocities, IvP/O (t0) = v∞e1 and IvQ/O (t0) = −v∞e1, we can use Eq. (7.40) to find
the initial velocities of each relative to G in C:
mQ − mP
C
vP/G(t0) = v∞e1 +
mP + m Q
v ∞ e1 =
mQ − mP
C
vQ/G(t0) = −v∞e1 +
mP + m Q
2mQ
mP + m Q
v ∞ e1 = −
v∞ e1
2mP
v∞e1.
mP + m Q
(7.46)
(7.47)
Using the geometry shown in Figure 7.11a and the expressions for the two velocities in Eqs. (7.46) and (7.47), the total angular momentum is
hG(t) = ChG(t0) =
C
2mP mQ
mP + m Q
=−
v∞ rP/G(t0) × e1 − rQ/G(t0) × e1
2mP mQ
mP + m Q
(7.48)
v∞be3.
We can also use the center-of-mass corollary to substitute for rQ/G and CvQ/G in the
total angular momentum expression in Eq. (7.45) to rewrite it in terms of the angular
momentum of P with respect to G only:
C
C
hG = mP rP/G × vP/G + mQ
=
=
mP + mQ
mQ
mP + mQ
mQ
mP
mQ
2
rP/G × CvP/G
mP rP/G × CvP/G
C
hP/G.
This equation states that the angular momentum of P with respect to G and the angular
momentum of Q with respect to G are both conserved.
ANGULAR MOMENTUM AND ENERGY OF A MULTIPARTICLE SYSTEM
283
Equating this expression for the total angular momentum about G with Eq. (7.48)
gives an expression for the specific angular momentum of P with respect to G in
terms of the initial speeds:
Ch
P/G
mP
=−
2m2Q
(mP + mQ
v∞be3 = hP e3.
)2
(7.49)
Eq. (7.49) allows us to perform the same substitution in the polar equations
of motion as we did in Example 4.9 for the specific angular momentum in polar
coordinates, hP = rP2 θ̇P , so Eqs. (7.43) and (7.44) become
θ̈P =
−2ṙP hP
rP3
(7.50)
r̈P =
h2P
μP
+
.
rP2
rP3
(7.51)
The result is the separation of the equations of motion so that we have a single
differential equation for rP , as in Eq. (4.47). We can thus follow the same procedure
as in Tutorial 4.2, except we now make the change of variables y = −1/rP to account
for the sign change due to the repulsive rather than attractive force. Substituting in
Eq. (7.51), we find
d 2y
μ
+ y = 2P .
hP
dθP2
This is the equation for simple harmonic motion, which, when put in terms of rP ,
gives the conic equation
rP (θP ) =
−h2P /μP
1 + e cos(θP − θ0)
,
where θ0 is the initial angle and e is the eccentricity.
Assuming that P starts at infinity, as in Figure 7.11, θ0 = π and the polar equation
becomes
h2P /μP
rP (θP ) =
> 0.
(7.52)
e cos θP − 1
The radial position rP in Eq. (7.52) is always positive since, as we’ll show next, the
eccentricity of the scattering encounter is greater than 1. The position of Q relative
to G is then found from the center-of-mass corollary, rQ/G = −(mP /mQ)rP/G.
Computing the total angular momentum about G allowed us to find the equation
for the orbit. To get the shape of the orbit we need the eccentricity e, which is found by
examining the total energy, as in Tutorial 5.2. For the closed orbits considered there,
the energy was negative and the eccentricity was thus less than one (corresponding to
circular or elliptical orbits). Here the total energy is positive because of the nonzero
velocity when the particles are at infinity. To see this, consider the constant total energy
of the system, using TG/O = 0,
1
1
EO = mP CvP/G2 + mQCvQ/G2 + UP/Q = constant,
2
2
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CHAPTER SEVEN
where UP/Q is the potential energy of the two particles (computed by integrating the
Coulomb force using rQ/P = ∞ as the zero reference),
UP/Q =
K
.
rQ/P Again using the center-of-mass corollary, we can rewrite the total energy as
EO =
KmQ 1
1 mP mP + mQ CvP/G2 −
.
m P + m Q rP
2 mQ
Following the same procedure as in Tutorial 5.2, we substitute the velocity in polar
coordinates to find the specific energy
εO =
mQ
h2
1
μ
1
EO = ṙP2 + P2 − P .
2
rP
m P + m Q mP
2rP
This equation is the same as Eq. (5.20). The eccentricity in terms of the specific energy
and angular momentum is the same as Eq. (5.22):
e=
1+
2εO h2P
μ2P
.
(7.53)
As noted, for closed orbits the energy is negative and thus the eccentricity is less
than one. Here the energy is positive, as seen by computing the total specific energy
at the initial time when the particles are at infinity, so that UP/Q = 0 and
2
2mQ
1
v 2 > 0.
εO = CvP/G(t0)2 =
2
(mP + mQ)2 ∞
Substituting into Eq. (7.53) gives the eccentricity,
4 b2
16m2P m2Qv∞
e = 1 + 2
> 1.
K (mP + mQ)2
The particles thus follow a hyperbolic trajectory as they scatter off each other. We
often describe such a hyperbola by the asymptotes or the straight-line trajectories at
infinity. The angle between the asymptotes of a hyperbola,
1
,
φ = 2 arcsin
e
is called the turning angle. It indicates the angle by which the trajectory has rotated
due to the scattering encounter.
Figure 7.12 shows an example of a pair of scattering trajectories. The two particles
have equal charge and equal mass, making the center of mass fixed at the origin of
the inertial frame.
ANGULAR MOMENTUM AND ENERGY OF A MULTIPARTICLE SYSTEM
285
20
15
Position of mQ (m)
10
5
0
ϕ
–5
–10
–15
–20
–6
–4
–2
0
2
Position of mP (m)
4
6
Figure 7.12 Trajectories of a scattering encounter between two particles of equal charge and
mass, traveling in opposite directions at 5 m/s in the inertial frame. The dashed line shows an
asymptote of the resulting hyperbolic trajectory. Turning angle φ is also shown.
7.6 Key Ideas
.
The total angular momentum of a collection of particles is equal to the sum of
the angular momentum of each individual particle:
I
hO =
N
mi ri/O × Ivi/O .
i=1
When the internal-moment assumption holds, the time derivative of IhO is equal
to the total external moment on the system:
Id
dt
.
.
I
hO =
N
(ext)
.
ri/O × Fi(ext) = MO
i=1
When the internal-moment assumption holds and the total external moment acting
on a system of particles about O is zero, then the total angular momentum of
the collection about O is conserved.
The total angular momentum of a collection of particles relative to point O can be
separated into the angular momentum of the center of mass about O, IhG/O , and
the angular momentum of the collection about the center of mass, IhG:
hO = IhG/O + IhG,
I
286
CHAPTER SEVEN
where
I
hG/O =
mGrG/O × IvG/O
and
I
hG =
N
mi ri/G × Ivi/G.
i=1
.
The dynamics of each of the angular momenta IhG/O and IhG are
Id
dt
I
(ext)
hG/O = MG/O
and
Id
dt
I
(ext)
hG = M G
,
(ext)
(ext)
is the external moment on the center mass relative to O and MG
where MG/O
is the external moment acting on all the particles relative to the center of mass.
.
For some problems, it is convenient to consider the angular momentum of all the
particles relative to an arbitrary point Q:
I
hQ =
N
ri/Q × mi Ivi/Q.
i=1
In terms of the angular momentum of the particles about their center of mass, this
angular momentum can be written as
I
hQ = IhG + mGrQ/G × IvQ/G.
.
The dynamics of the angular momentum IhQ are
Id
dt
(ext)
hQ = MQ
+ rQ/G × mGIaQ/O .
I
The equation of motion for rotation relative to Q is thus coupled to the translational
motion of Q.
.
The total kinetic energy of a collection of particles is equal to the sum of the
kinetic energy of each individual particle:
N
N
1 I
1
I
.
Ti/O =
mi vi/O vi/O =
mi Ivi/O 2.
TO =
2
2
i=1
i=1
i=1
.
N
As with the total angular momentum, the total kinetic energy of the particles
separates into the kinetic energy TG/O of the center of mass and the kinetic energy
ANGULAR MOMENTUM AND ENERGY OF A MULTIPARTICLE SYSTEM
287
TG of the particles relative to the center of mass:
TO = TG/O + TG,
where
1
mGIvG/O 2
TG/O =
2
and
TG =
.
N
1
mi Ivi/G2.
2 i=1
The total work on a system of particles is equal to the sum of the external and
internal work:
W (tot) = W (ext) + W (int).
.
The power on the center of mass of the particles is related to the total external
force acting on the center of mass and the change in kinetic energy of the center
of mass:
(ext) . I
vG/O (t) =
PG = FG
d
TG/O (t).
dt
If the total external force is zero, the kinetic energy of the center of mass is a
constant of the motion.
.
Using E =
E(t2) − E(t1), the work-energy formula for a multiparticle system
is
E = TG/O + TG + UO(ext) + UO(int)
(nc,int)
({ri/O ; γi }N
= W (nc,ext)({ri/O ; γi }N
i=1) + W
i=1).
If the total nonconservative internal and external forces are zero, the total energy
of the system of particles is a constant of the motion.
7.7 Notes and Further Reading
To obtain one of the main ideas of this chapter, Eq. (7.5), which equates the rate
of change of total angular momentum to the external moment on the collection, we
had to make an assumption about the internal moments acting among the particles.
Remarkably, a surprising number of texts pass over this issue quite quickly. They
either ignore the internal-moment term in Eq. (7.3) (e.g., Greenwood 1988; Tongue
and Sheppard 2005), incorrectly state that Newton’s third law implies that the internal
forces act along the line connecting the particles (e.g., Beer et al. 2007; Ginsberg
2008), or invoke Newton’s third law to state that the sum of the internal moments
in Eq. (7.3) must be zero (e.g., Meriam and Kraige 2001; Rao 2006). We have
288
CHAPTER SEVEN
emphasized that Eq. (7.5) only holds for those cases where the internal-moment
assumption applies (and we provide a counterexample in Problem 7.13). This subtle
and historically contentious point dates back to Bernoulli and Euler and reminds us
of the importance of always checking the assumptions that go into solving a problem.
An excellent discussion of the history of the “law of moment of momentum” can be
found in the engaging book of essays by Truesdell (1968). We revisit this point again
in the notes for Chapter 9.
We also emphasized the importance of considering the internal forces and their
contributions to the internal work on a system of particles. As in the derivation of the
angular momentum equation, some texts ignore the internal forces when computing
the work and energy of a collection. Among those that do include internal work is
Bedford and Fowler (2002), where we found the material for Examples 7.9 and 7.10.
7.8 Problems
7.1
Consider the three-particle system shown in Figure 7.13. The mass, position,
and velocity of each particle are m1 = m, r1/O = ex + 2ey , Iv1/O = −ex − ey ;
m2 = 2m, r2/O = 2.5ex + ey , Iv2/O = 2ey ; and m3 = 3m, r3/O = ex + ey ,
Iv
3/O = ex .
I
ey
m
I
I
v2/O
v1/O
3m
I
v3/O
2m
O
ex
Figure 7.13 Problem 7.1.
a. Find the total angular momentum IhO about O.
b. Find the angular momentum IhG/O of the center of mass about O.
[HINT: First find the center of mass rG/O .]
c. Find the angular momentum IhG about the center of mass.
d. Show that IhO = IhG/O + IhG.
7.2
Repeat Problem 7.1 for kinetic energy:
a.
b.
c.
d.
Find the total kinetic energy TO with respect to O.
Find the kinetic energy TG/O of the center of mass about O.
Find the kinetic energy TG about the center of mass.
Show that TO = TG/O + TG.
ANGULAR MOMENTUM AND ENERGY OF A MULTIPARTICLE SYSTEM
7.3
289
Find the equation of motion for the seesaw shown in Figure 7.14.
l
2m
m
2l
g
Figure 7.14 Problem 7.3.
7.4
A particle of mass m1 slides without friction along a circular track of radius R
cut from a block of mass m2, as shown in Figure 7.15. The block slides without
friction along a horizontal surface. Assume that x describes the horizontal
displacement of the block and that θ describes the position of the particle
relative to the vertical. Assume that x(0) = 0, ẋ(0) = 0, and θ̇(0) = 0. Be
sure to explicitly point out any use of conserved quantities, such as energy or
momentum.
m2
g
R
m1
θ
Figure 7.15 Problem 7.4.
a. If the particle starts at θ(0) = θ0, what is the velocity of the particle
and the velocity of the block when θ = 0? [HINT: Start by carefully
formulating the inertial velocity of the block and the particle in terms
of x, ẋ, θ, and θ̇.]
b. If friction is introduced so that the particle oscillates until it lies at
rest at θ = 0, what is the steady-state position of the block, assuming
θ(0) = π/2?
7.5
The baton shown in Figure 7.16 is leaning against a vertical wall. Because
there is no friction between the baton and the wall or between the baton and
the floor, the baton slides along the floor. Find the angle θ when the baton
falls. [HINT: The baton falls when the normal force from the wall is zero.]
I
m
θ
g
l
m
O
Figure 7.16 Problem 7.5.
290
CHAPTER SEVEN
7.6
Consider the two masses connected by a spring and damper from Example 7.11. Suppose they start moving with the initial conditions shown in
Figure 7.17, with particle P moving to the right at a speed v0. After reaching
its steady-state motion, the spring has stretched to twice its original length.
Assuming that the two particles have equal mass m, show
√that v0 as a function of the other system parameters is given by v0 = 4r0 k/m, where r0 is
the unstretched length of the spring and k is its spring constant. Compute the
work done by the damper as a fraction of the total initial energy.
Q
v0
P
Figure 7.17 Problem 7.6.
7.7
Three identical masses lie on a frictionless horizontal surface. One mass is
at rest and the other two masses, which are connected by a string of length
l, slide with initial velocity v0, as shown in Figure 7.18a. The middle of the
string meets the stationary mass, and all three masses slide together, as shown
in Figure 7.18b. When the two masses at each end of the string collide, they
stick together (Figure 7.18c).
m
m
l
v0
1/2
1/2
m
vf
(a)
(b)
(c)
Figure 7.18 Problem 7.7. (a) Initial setup. (b) Intermediate configuration. (c) Final configuration.
a. Find the final velocity vf of the three-mass system after the collision.
[HINT: There are no external horizontal forces.]
b. Find the velocity of each of the three masses at the instant of the
collision.
7.8
Consider again the colliding rail cars in Problem 6.14. What is the maximum
compression of the spring after the collision?
ANGULAR MOMENTUM AND ENERGY OF A MULTIPARTICLE SYSTEM
vA = 900 m/s
291
B
50 g
A
C
30°
Figure 7.19 Problem 7.9.
7.9
The 10 kg wedge B shown in Figure 7.19 is held at rest on an incline of 30◦ by
stop C when it is struck by the 50 g bullet A, which is traveling horizontally
with speed vA = 900 m/s. Assume that the bullet is stopped instantaneously
by the block.
a. Calculate the velocity with which the wedge starts up the incline.
b. Assuming gravity acts vertically downward, how high does the block
travel before stopping? (Assume zero friction.)
7.10 Suppose an apple of mass M is hanging motionless on a massless string of
length l from a fixed point O when it is impaled by an arrow of mass m moving
horizontally with speed v0 (Figure 7.20). What is the maximum angle θ that
the apple reaches after the collision?
O
g
θ
l
v0
Figure 7.20 Problem 7.10.
7.11 Consider two masses m1 and m2 interacting gravitationally with no external
forces, as in Tutorial 7.1. Find the equations of motion of each mass relative
to the center of mass of the system. What do you think the resulting orbits
look like?
7.12 James Bond is being chased by a group of S.P.E.C.T.R.E. agents and spots a
ramp in front of a canyon. In an attempt to escape, he tries to jump his Aston
Martin DB5 across the canyon. The ramp has an angle α and the canyon is L
meters wide. Bond has mass mP and the Aston Martin has mass mC .
a. Given that Bond drives off the ramp at constant speed vo , what is the
greatest possible width L the canyon can have so that Bond escapes
safely?
b. Find an expression for Bond’s horizontal position, relative to the edge
of the ramp, when he reaches the apex of the trajectory. What must this
equal so that Bond knows he can cross the canyon safely at this point?
c. At the peak of the trajectory Bond has either decided he won’t make it,
or he is simply taking some time out of his day to irritate Q by returning
the car in less than pristine order. Regardless, he makes the decision
to use the ejector seat. Assume that Bond and the car are two point
292
CHAPTER SEVEN
v0
L
α
Figure 7.21 Problem 7.12.
masses. Model the ejector with a spring of constant k compressed by
ξ . Assume that the spring is released quickly, so the resulting force
is well modeled by an impulse. Also assume that the impulse due to
the spring acts purely in the vertical direction between mass mP and
mC . What is Bond’s velocity IvP/O (t ∗) immediately after deployment
(where t ∗ is the time he reaches the apex)?
d. Suppose that the canyon length is indeed too long to cross without
ejection. Solve for the spring compression ξ of Bond’s ejector seat
that ensures a safe crossing of the canyon. Express your answer in
terms of L, v0, α, k, mP , and mC .
7.13 Consider a slightly different scattering experiment than the one described in
Tutorial 7.3. Rather than two charged particles interacting, here you analyze
the encounter of a positively charged particle Q with a magnetic dipole P .
Assume Q has charge +q and P has a dipole moment m = me3, so that the
dipole moment is always perpendicular to the plane of the encounter. The
magnetic field B due to a dipole is
B=
μ0 . r̂)r̂ − m ,
3(m
4πr 3
where r is the distance from the dipole, μ0 is the permeability of free space,
and r̂ is the unit vector in the direction of the dipole. Assume that both particles
have mass mQ and initial velocities IvQ/O (t0) and IvP/O (t0). The force on Q
due to P is given by the force due to a dipole,
FQ,P = q IvQ/O × B,
since we have confined the dipole to be perpendicular to the plane. By Newton’s third law, the force on P must be equal and opposite.
a. What is the velocity of the center of mass? Is it conserved?
b. Find an expression for the total angular momentum of the two particles,
relative to an inertially fixed point. Is it conserved? Is the motion still
an orbit in the plane?
c. Find the equations of motion for the two particles relative to a frame
with origin you chose in (b).
d. Simulate the trajectories of the two particles, and plot on the same set
of axes. Also plot the angular momentum you computed in (a). Does it
confirm your conclusion in (b)? Does the internal-moment assumption
hold for this system? Assume the particles each have a mass of 1 kg, an
impact parameter b of 0.1 m, and equal and opposite initial velocities
relative to their center of mass of 5 m/s. Q has a charge of 1 C and the
magnetic dipole moment of P is 105 A-m2. In these units, μ0 = 1.
PART THREE
Relative Motion and Rigid-Body
Dynamics in Two Dimensions
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CHAPTER EIGHT
Relative Motion in a Rotating Frame
This chapter is devoted to relative motion in a rotating reference frame. Of course,
you have already been introduced to some rotating frames, namely, the polar and
path frames. These frames were enormously useful for solving certain problems,
simplifying the kinematics, and providing new insights into particle motion. This
chapter examines problems where using multiple rotating frames leads to important
new formulas for dynamical systems. In particular, we see how the kinematics of
certain planar rigid bodies are simplified by attaching a reference frame to them and
then applying our previous results. We then complete the study of planar kinematics
by examining the relative motion of a particle in a translating and rotating frame of
reference. This expands our treatment in Chapter 3, where we restricted the discussion
of relative motion to a translating frame. This chapter introduces one of the most
important formulas of the book and the precursor to our subsequent treatment of
rigid-body dynamics—the transport equation.
8.1 Rotational Motion of a Planar Rigid Body
This section focuses on the kinematics of a rotating and translating rigid body. We
have alluded to rigid bodies before without rigorously defining them. For now, you
simply need to recognize that a rigid body is a collection of particles constrained so as
not to move relative to one another. In many cases we consider a continuous collection
that forms a solid body, such as a disk, rod, or sphere. Alternatively, we may consider
a finite collection of particles connected by massless rigid links. What is important
for our study of kinematics is that to every rigid body we can attach a body frame.
Recall Definition 3.1: a reference frame is equivalent to a rigid body. Since all points
of a rigid body are fixed with respect to one another, we can use them to define a
reference frame. Thus as a rigid body translates and rotates in inertial space, so will a
body frame attached to it. The distinction between the polar frame and the body frame
here is subtle: a polar frame always points toward a moving point, whereas a body
296
CHAPTER EIGHT
B
b2
I
P
rP/O′
Ꮾ
e2
b1
O′
rO′/O
O
e1
Figure 8.1 Planar rigid body B with attached body frame B translating and rotating in the
plane with angular velocity IωB relative to inertial frame I .
frame is fixed to a rigid body. As shown later, in many ways the body frame behaves
just like a polar frame. This observation will allow us to reuse many previous results.
Figure 8.1 shows a rigid body B with an attached body frame B = (O , b1, b2, b3),
where the origin O is a point on the rigid body. O is not necessarily the center of mass.
The position of O with respect to O is rO /O . The rigid body (and thus frame B) is
rotating relative to the inertial frame with an angular velocity IωB , as well as possibly
translating with a velocity IvO /O and acceleration IaO /O . Point P is an arbitrary point
fixed to the rigid body, which implies that P does not move in frame B.
Suppose we want to find the velocity of point P in the inertial frame. We start by
writing the position of P relative to O using the vector triad,
rP/O = rO /O + rP/O .
We find the velocity of P by taking the time derivative of rP/O :
I
vP/O =
Id rP/O .
rO /O +
dt dt
Id
(8.1)
=IvO /O
The first term on the right-hand side of Eq. (8.1) is the translational velocity of
B, that is, the velocity IvO /O of the origin of frame B. What about the second term?
Although P has no velocity in B, it is moving in I due to the rotation/translation of
the body.
We approach this question the same way we approached the velocity in the polar
frame in Chapter 3. Writing out the position of P as components in the body frame
using the Cartesian coordinates (x, y)B leads to
I
vP/O =
Id
rP/O = (xb1 + yb2)
dt
dt
Id
=x
Id
dt
b1 + y
Id
dt
b2 ,
(8.2)
RELATIVE MOTION IN A ROTATING FRAME
297
where we have used the distributive property and product rule of vector differentiation
and the fact that P is fixed in B. The unit vector derivatives in Eq. (8.2) are the same as
those treated in Chapter 3 when studying the polar and path reference frames. Since
I B
ω b3, the unit-vector
B is rotating with respect to I with angular velocity IωB =
derivatives are
Id
dt
bi = IωB × bi
i = {1, 2}.
(8.3)
This is a good time to review Section 3.4.2. We are still restricting the motion to be
planar, so the rotation of frame B with respect to I means a simple rotation by some
angle θ about the e3 or, equivalently, b3 axis. The angular velocity is thus a simple
angular velocity and is given by IωB = θ̇b3. Note that the rotation rate θ̇ need not be
constant! Although in Section 3.4.2 the development of Eq. (8.3) was for the polar
(or path) frame, the derivation was completely general. Eq. (8.3) describes the time
derivative of an arbitrary unit vector bi that is fixed in a frame B rotating with respect
to frame I with angular velocity IωB .
Now we can substitute Eq. (8.3) into Eq. (8.2) to find
I
vP/O = x(IωB × b1) + y(IωB × b2)
or, after using the distributive rule of the cross product,
vP/O = IωB × (xb1 + yb2) .
I
(8.4)
=rP/O
The last term is the position of P in B. This identification lets us write the final
expression for the velocity of P in I by combining Eqs. (8.1) and (8.4):
vP/O = IvO /O + IωB × rP/O .
I
(8.5)
Eq. (8.5) is a fundamental and important equation. It is our basic tool for finding
the velocity of a point on a rigid body undergoing translational and rotational motion.
Here we derived it under the restriction of planar rotation of the rigid body; later chapters show that it is completely general. It is the starting point for many applications
in mechanics, some of which are shown in the following examples.
Example 8.1 Kinematics of a Compound Pendulum
A compound pendulum is a generalization of the simple pendulum we have been
repeatedly studying. Rather than a particle attached to the end of a massless rod, a
compound pendulum is an extended rigid body pinned at some point and allowed to
swing, as shown in Figure 8.2. We restrict ourselves to planar motion, so this system
has a single degree of freedom—the angle of the pendulum. We use the coordinate θ
to measure the angle from the local vertical to a line connecting the pin to the center of
mass. Figure 8.2b shows inertial frame I = (O, e1, e2, e3) located at the pin, labeled
O, and body frame B = (O, b1, b2, b3) fixed to the pendulum at O. The angular
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CHAPTER EIGHT
b2
B
e2
O
I
O
θ
l
e1
b1
G
G
(a) Coordinates
(b) Reference frames
Figure 8.2 Compound pendulum consisting of a rigid body pinned at point O in the inertial
frame.
velocity of the pendulum is thus IωB = θ̇b3. In this example we calculate the inertial
velocity IvG/O of the center of mass of the pendulum.
Because the pendulum is pinned at O, there is no translational velocity (IvO /O =
0), and the origin of B is the origin of I . Using Eq. (8.5), the velocity of the center
of mass is
I
vG/O = IωB × rG/O .
Writing the vectors in the body frame, we obtain
I
vG/O = θ̇b3 × lb1 = l θ̇b2.
Example 8.2 The Velocity of a Rolling Wheel
This example computes the velocity of the center of a rolling wheel or disk, shown
in Figure 8.3a. The rolling disk is an essential construct of mechanics and is used as
one of the elements for building a dynamic model of many systems. The key idea is
that the disk rolls without slipping. That means the disk actually only has one degree
of freedom, even though it is both translating and rotating. Thus there must be a
constraint. We call the constraint the no-slip condition. The no-slip condition states
that the point of the disk making contact with the ground has no instantaneous inertial
velocity (that is, it has the same velocity as the point it is touching on the ground). We
use this constraint along with Eq. (8.5) to find the translational velocity of the center
of the wheel.
Figure 8.3b shows the frames used for this problem. The no-slip condition states
that IvP/O = 0, where P is the point of contact with the ground. In this case, Eq. (8.5)
becomes
0 = IvO /O + IωB × rP/O ,
where IωB = ωB b3, as drawn in Figure 8.3 (a positive rotation—the disk rolls to the
left). Since the position of P relative to the center of the wheel is rP/O = −Rb2, we
have
I
vO /O = −ωB b3 × (−Rb2) = −RωB b1.
RELATIVE MOTION IN A ROTATING FRAME
299
B
I
ωB
b2
e2
ωB
R
rO′/O
O
b1
O′
rP/O′
P
e1
(a)
(b)
Figure 8.3 A disk of radius R rolling without slipping on a horizontal surface at angular rate
ωB . (a) Rolling disk. (b) Reference frames.
In Cartesian coordinates, where rO /O = xe1 + Re2, the no-slip condition is
ẋ = −RωB .
(8.6)
The no-slip condition in Eq. (8.6) is an example of a motion constraint, that
is, a constraint on speeds rather than on coordinates. A motion constraint reduces
the number of degrees of freedom in a problem but not necessarily the number of
coordinates in terms of the others (see Section 2.4). In the case of the no-slip condition,
Eq. (8.6) can be integrated if we know the initial position or angle of the wheel. Thus,
we only need one coordinate for the one degree of freedom. We discuss general motion
constraints in Chapter 13.
Example 8.3 The Gear Ratio
As you may know from riding a bike, gears are used to change an angular velocity.
For example, a motor may produce an angular velocity that is faster than desired and a
gear is used to reduce it. The gear ratio dictates the amount of reduction (or increase).
Figure 8.4a shows two gears with centers fixed in inertial space. Gear A has radius
rA and is rotating at rate ωA. If gear B has radius rB , what is its angular rate ωB ?
Like the rolling disk discussed in the previous example, this problem has only one
degree of freedom. That is, once the angle of gear A is set, the angle of gear B is
b1
B
I
e2
rB
b2
ωB
ωB
O′
rP/O′
A
P
a2
Ꮾ
rA
rP/O″
Ꮽ
O
ωA
(a)
O″
a1
ωA
e1
(b)
Figure 8.4 (a) Two fixed, interlocking gears of radius rA and rB . Gear A is rotating in the
plane at the angular rate ωA, and gear B is rotating at ωB . (b) Reference frames.
300
CHAPTER EIGHT
determined. Since there are two coordinates (the angles of each gear), there must be a
constraint. We call this constraint the gear constraint.The gear constraint requires that
the points of contact between the two gears have the same inertial velocity. Otherwise,
the gears would slip.
We solve this problem again using Eq. (8.5). Figure 8.4b shows the body frames
A and B fixed to each gear. Point P is the point of contact between the two gears.
Since the gears are fixed, the velocities of their centers are zero:
I
vO /O = IvO
/O
= 0.
Using Eq. (8.5) on A to find the velocity of P , we have
I
vP/O = IωA × rP/O ,
where IωA = −ωAa3. Likewise, for B we have
I
vP/O = IωB × rP/O ,
where IωB = ωB b3. By the gear constraint, these two velocities must be the same:
ωB b3 × (rB r̂P/O ) = ωAa3 × (rAr̂P/O ) = ωAb3 × (−rAr̂P/O )
which implies
ωB rB = −ωArA.
Eq. (8.7) yields the following equation, called the gear equation:
r
ω B = − A ωA .
rB
(8.7)
(8.8)
The ratio rA/rB is the gear ratio. Eq. (8.8) is another example of a motion constraint.
Example 8.4 A Two-Bar Linkage
A linkage consisting of two or more long, rigid bars is a device that converts rotational
motion into translational motion or transfers rotational motion from one location to
another (e.g., from the steering wheel of a car to the front wheels). In a three-link robot
arm, for example, each link has one degree of freedom; the coordinate associated
with each degree of freedom is given by the angle of the link (e.g., in the robot arm
of Example 2.5, there are three angles: the shoulder, the elbow, and the wrist).
Linkages are ubiquitous in mechanical systems, yet all linkages rely on the same
principle. If the rotation rate of each joint is known, the velocity of a point on the
linkage can be found using Eq. (8.5). Usually linkages are combined with mechanical
constraints to perform specific types of motion.
Figure 8.5 shows a simple two-bar linkage. This system has only one degree of
freedom because the second link is constrained to always slide (without friction) on
the horizontal surface. If the angular velocity of bar A is given as IωA = ωAa3, what
is the angular velocity ωB b3 of bar B, and what is the velocity of its endpoint P ?
Using Eq. (8.5), the velocity of P is
I
vP/O = IvO /O + IωB × rP/O .
RELATIVE MOTION IN A ROTATING FRAME
301
O′
ωA
0.4 m
Ꮽ
Ꮾ
P
O
0.4 m
0.8 m
(a)
B
b2
I
e2
O′
b1
a1
a2
A
ωA
e1
O
P
(b)
Figure 8.5 (a) Two-bar linkage constrained to move along a horizontal surface. (b) Reference
frames.
Here the velocity IvO /O is also due to a rigid body rotation and is thus given by the
same formula in reference frame A:
vO /O = IωA × rO /O .
I
Combining these two expressions yields
vP/O = IωA × rO /O + IωB × rP/O .
I
From Figure 8.5a we can write down the instantaneous positions of P and O in I ,
rO /O = 0.4e1 + 0.4e2
rP/O = 0.8e1 − 0.4e2.
Substituting for the positions as components in I gives the following expression
for the velocity of point P :
I
vP/O = ωAa3 × (0.4e1 + 0.4e2) + ωB b3 × (0.8e1 − 0.4e2)
= 0.4ωAe2 − 0.4ωAe1 + 0.8ωB e2 + 0.4ωB e1
= (−0.4ωA + 0.4ωB )e1 + (0.4ωA + 0.8ωB )e2.
(8.9)
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CHAPTER EIGHT
Now we use the constraint that point P moves only on the horizontal plane, which
means its velocity is in the e1 direction:
I
vP/O = vP e1.
We can now solve for the two unknowns, vP and ωB . Since the e2 component in
Eq. (8.9) is zero, we have
0 = 0.4ωA + 0.8ωB ,
which results in
ωB = −
ωA
rad/s.
2
The e1 component yields the speed of point P as a function of ωA:
vP = −0.4ωA + 0.4ωB = −0.4ωA − 0.2ωA = −0.6ωA m/s.
8.2 Relative Motion in a Rotating Frame
Section 3.6 introduced the idea of relative motion, though restricted to translation
only. We asked there how to develop the equations of motion of a particle in terms of
a set of coordinates defined relative to a moving frame of reference. Since Newton’s
second law applies only to accelerations relative to an inertial frame, we had to be
careful with the kinematics. We found a correction term to account for the translational
acceleration of the frame. Now we generalize the treatment of relative motion to a
frame that is both translating and rotating and develop the kinematics of a particle
moving in a rotating frame. We then show how to use Newton’s second law to develop
the dynamics of a particle relative to a rotating frame. This problem is exceptionally
important. For instance, every dynamics problem we solve on the surface of the earth
is, in fact, relative to a translating and rotating frame!
8.2.1 The Transport Equation
Section 8.1 treated the problem of finding the inertial velocity of a particle P fixed to
a translating and rotating frame of reference (a rigid body). In this section, we relax
the restriction that P be fixed to the body and treat the more general problem of P
moving relative to the body frame. In other words, suppose we know both the position
and velocity of a particle in a reference frame B that is translating and rotating in the
inertial frame. What is its inertial velocity? It turns out that answering this question
requires only a slight extension of our earlier treatment.
The situation is illustrated in Figure 8.6, where body frame B = (O , b1, b2, b3)
is translating and rotating relative to inertial frame I = (O, e1, e2, e3). Note the
similarity to Figure 3.26. The important difference is that we now allow the body
frame to also rotate in the plane. Particle P is free to move in the plane at any velocity
relative to frame B.
RELATIVE MOTION IN A ROTATING FRAME
I
303
P
B
e2
b2
rP/O′
rP/O
b1
rO′/O
O′
I
vO′/O
O
e1
Figure 8.6 Body frame B translating and rotating with respect to inertial frame I .
To find the inertial velocity of P , we begin as in Section 8.1, by taking the derivative
of the vector triad locating P relative to O:
I
vP/O =
Id rO /O +
rP/O = IvO /O + IvP/O .
dt
dt
Id
(8.10)
The first quantity on the right, IvO /O , is the translational velocity of frame B.
The second quantity is the inertial velocity of P with respect to O . We write the
components of rP/O using Cartesian coordinates (x,y)B and take the derivative:
I
vP/O =
Id
Id
(xb1 + yb2) = ẋb1 + ẏb2 +x b1 + y b2.
dt
dt
dt
Id
(8.11)
=BvP/O Since P can move in B, we must include the rates of change of the Cartesian
coordinates. This introduces the relative velocity of P with respect to frame B, BvP/O .
The last two terms in Eq. (8.11) result from the rotation of B in I . We already solved
for these two terms in the previous section. Eq. (8.11) thus simplifies to
I
vP/O = BvP/O + IωB × rP/O ,
(8.12)
or, explicitly writing the derivatives,
Bd rP/O + IωB × rP/O .
rP/O =
dt
dt
Id
(8.13)
If B is not rotating, then IωB = 0 and Eq. (8.12) reproduces what we showed in
Section 3.6, that is, IvP/O = BvP/O . However, the rotation of B adds the correction
term IωB × rP/O . Eq. (8.12) is the most general equation for relating the inertial
velocity of a particle to its motion in a translating and rotating frame.
Eq. (8.13) is an extraordinarily important equation, perhaps the most important one
we encounter (short of Newton’s second law). In fact, what is key about Eq. (8.13)
is that it holds for any vector. It is not restricted to just the position. Thus the most
general form of Eq. (8.13) is written in terms of an arbitrary vector c and two arbitrary
reference frames A and B. Since we can decompose c into components in the unitvector directions of frame B and write the rotation of B in A using the angular
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velocity AωB , the above development holds without modification. We call the resulting
equation the transport equation. It is the backbone of all the kinematics to follow.
Memorize it!
Definition 8.1 The transport equation is used to find the time derivative with
respect to frame A of a vector c in terms of the time derivative of c with respect
to frame B and the angular velocity AωB of B relative to A:
Ad
dt
c=
Bd
dt
c + AωB × c.
Note that we can write the vectors on either side of the transport equation as
components in either frame A or frame B, whichever is most convenient for the
problem at hand. We use Eqs. (8.10) and (8.12) to obtain the final expression for the
inertial velocity of point P , given the translational velocity of frame B, the angular
velocity of B with respect to I , and the relative velocity of P in B:
I
vP/O = IvO /O + BvP/O + IωB × rP/O .
(8.14)
Example 8.5 Throwing the Brass Ring
You may have had the pleasure of riding a carousel, grabbing a brass ring, and trying
to throw it into a clown’s mouth while the carousel is rotating. The challenge is to
figure out just the right speed and direction to throw the ring relative to the carousel.
We assume that no (horizontal) forces are acting on the ring once it leaves your hand
so that it travels in a straight line in an inertial frame with whatever initial velocity
you give it.
A schematic of the situation is shown in Figure 8.7. The objective is for the ring to
have the velocity IvP/O in the inertial frame so that it will land in the clown’s mouth.
You are located at point O on the carousel, which is rotating with angular velocity
IωB = ω e (frame B is fixed to the carousel). We will find the velocity Bv
B 3
P/O that
I
corresponds to vP/O = vP r̂C/O .
Solving Eq. (8.14) for the body-frame velocity yields
vP/O = IvP/O − IvO /O − IωB × rP/O .
B
At the instant you throw the ring it is at O , which implies rP/O = 0. Since the carousel
is going in a circle, you can use the transport equation to find the velocity of your
horse (O ):
vO /O = BvO /O +IωB × rO /O .
I
=0
Combining these two equations gives the desired velocity,
B
vP/O = IvP/O − ωB e3 × rO /O = vP r̂C/O − ωB e3 × rO /O .
(8.15)
This equation shows what makes the game fun: in a split second you must estimate
the angular velocity of the carousel, the position of your horse, and the direction to
the clown. You then have to solve Eq. (8.15) (in your head) and throw.
RELATIVE MOTION IN A ROTATING FRAME
305
C
I
B
vP/O
B
b2
I
e2
vP/O′
b1
O′
ωB
O
(a)
e1
(b)
Figure 8.7 (a) Carousel with a body frame B rotating in inertial frame I . (b) Carousel schematic
(top view). IvP/O is the velocity of a ring thrown at clown C.
Example 8.6 Linkage with a Sliding Contact
This example examines a linkage with a sliding contact, as shown in Figure 8.8. Unlike
the linkages in Example 8.4, here the two links are not connected at a pinned joint,
but rather a pin on one link is free to slide in a slot on the other link. This is similar to
the transfer of translational motion to rotational motion in a steam engine. The goal
is to find the angular velocity of the vertical bar and the velocity of the pin in the slot,
given a fixed angular velocity of the slotted link.
As in Example 8.4, we begin by defining an inertial frame fixed at O and attaching
body frames A and B to the two bars. We give bar A an angular velocity IωA = ωAa3
of 2 rad/s and ask for the angular velocity of bar B, IωB = ωB b3, and the inertial
velocity IvP/O of pin P in the slot at the instant shown in Figure 8.8.
P
P
Ꮽ
0.4 m
O
ωA
Ꮾ
A
a2
I
O′
O
0.3 m
(a)
e2
a1
B
θ
b2
O′
b1
e1
(b)
Figure 8.8 (a) A sliding linkage, where pin P is constrained to move along a slot in bar A .
(b) Reference frames.
306
CHAPTER EIGHT
We use the transport equation to write the inertial velocity of point P relative to
O in terms of the body-frame velocity AvP/O :
vP/O = AvP/O + IωA × rP/O .
I
(8.16)
Since rP/O = 0.3e1 + 0.4e2 and IωA = ωAe3 = 2e3 rad/s, Eq. (8.16) simplifies to
I
vP/O = AvP/O + 0.6e2 − 0.8e1.
(8.17)
Because of the slot, the velocity of P in A is always along a1, which implies
A
vP/O = vP a1 = vP (cos θe1 + sin θe2),
where we have used the transformation array between frames I and A,
e1
a1
a2
e2
cos θ
− sin θ
sin θ
cos θ ,
and θ = arctan(0.4/0.3) = 53.1◦.
Substituting this result back into Eq. (8.17) yields the inertial velocity of P with
respect to O as a function of vP :
I
vP/O = (vP cos θ − 0.8)e1 + (vP sin θ + 0.6)e2.
A second equation for IvP/O is found by examining the velocity of P in reference
frame B and again using the transport equation. As in Example 8.4, the velocity in B
is zero (as the pin is fixed to the bar), so the transport equation becomes
I
vP/O = IvO /O + BvP/O +IωB × rP/O =0
=0
= (ωB e3) × (0.4e2) = −0.4ωB e1.
We now have two equations for IvP/O that can be solved for vP and ωB . We have
(vP cos θ − 0.8)e1 + (vP sin θ + 0.6)e2 = −0.4ωB e1.
Equating components in the e1 and e2 directions yields the following two scalar
equations in the unknowns vP and ωB :
vP cos θ − 0.8 = −0.4ωB
vP sin θ + 0.6 = 0,
which implies
vP = −0.75 m/s
ωB = 3.125 rad/s.
RELATIVE MOTION IN A ROTATING FRAME
307
C
I
A
B
e2
b2
b1
O″
rO″/O′
O′
a2
a1
O
c2
c1
rP/O″
P
rO′/O
rP/O
e1
Figure 8.9 Multiple rotating body frames.
8.2.2 The Addition of Angular Velocities
Example 2.5 introduced a three-link robot arm as an example of a three-degree-offreedom system. Figure 2.6b shows a simple planar model of the arm, where the three
scalar angles of each arm relative to the previous one are used to describe its state.
This is a sensible set of coordinates, since motors and sensors are often located at the
joints and operate based on the relative rate between the two bodies. The question
is, how do we combine the angular velocities of multiple frames to find the angular
velocity with respect to an inertial frame? If we are going to study the dynamics of an
individual arm, we need to know its inertial angular velocity. In this section we study
how to go back and forth between the inertial and the relative angular velocities.
Figure 8.9 shows an arrangement of three connected frames, A = (O, a1, a2, a3),
B = (O , b1, b2, b3), and C = (O , c1, c2, c3). Each has some angular velocity, IωA ,
IωB , and IωC , respectively, in the inertial frame. Suppose we want to find the inertial
velocity of point P . We can use Eq. (8.14) to find
Id rP/O =
rO /O + CvP/O + IωC × rP/O .
dt
dt
Id
We can use the vector triad and the transport equation on the first term to rewrite
the derivative in terms of the translation and rotation of frame B:
Id rP/O =
rO /O + BvO /O + IωB × rO /O + CvP/O + IωC × rP/O .
dt
dt
Id
Finally, using Eq. (8.14) to replace the inertial velocity of point O in terms of the
angular velocity of reference frame A (which, in this setting, shares an origin with
the inertial frame) yields
I
vP/O = AvO /O + IωA × rO /O + BvO /O + IωB × rO /O + CvP/O + IωC × rP/O .
(8.18)
This equation shows how to use the transport equation on multiple frames of
reference to find the velocity of a point. However, as mentioned at the beginning
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CHAPTER EIGHT
of the section, we may not know the angular velocities with respect to an inertial
frame but rather with respect to an intermediate frame that may also be rotating. We
therefore rederive the velocity of P using a slightly different approach.
Rather than starting with the velocity of P in frame C and then working backward
through frames B and A, as we did above, we begin with the position of P as a vector
sum, take the derivative, and work forward. We have
rP/O = rO /O + rO /O + rP/O .
The inertial velocity is found by taking the vector derivative to obtain
I
vP/O =
Id Id rO /O +
rO /O +
rP/O .
dt
dt
dt
Id Now we use the transport equation separately on each term. We begin by transforming
to derivatives in frame A, using the angular velocity IωA of A relative to I (remember
that, in this example, A is not translating in I ):
I
vP/O = AvO /O + IωA × rO /O +
+
Ad dt
rO /O + IωA × rO /O rP/O + IωA × rP/O .
Ad dt
Next, we transform from derivatives in A to derivatives in B, using the angular
velocity AωB of B relative to A:
vP/O = AvO /O + IωA × rO /O + BvO /O + AωB × rO /O I
+ IωA × rO /O +
Bd dt
rP/O + AωB × rP/O + IωA × rP/O .
Finally, we use the transport equation one more time to get the velocity of P in frame
C, using the angular velocity BωC of C relative to B:
vP/O = AvO /O + IωA × rO /O + BvO /O + AωB × rO /O + IωA × rO /O I
+ CvP/O + BωC × rP/O + AωB × rP/O + IωA × rP/O .
This expression is a bit cluttered, so we simplify it by factoring and combining the
cross products:
vP/O = AvO /O + IωA × rO /O + BvO /O + IωA + AωB × rO /O (8.19)
+ CvP/O + IωA + AωB + BωC × rP/O .
I
Compare Eq. (8.19) to Eq. (8.18). Both are expressions for the inertial velocity
of P , IvP/O . The two expressions must be the same. Comparing each term reveals
RELATIVE MOTION IN A ROTATING FRAME
309
an extremely important property of angular velocities, known as the angular-velocity
addition property:
I B
ω = IωA + AωB
and
I C
ω = IωA + AωB + BωC .
In fact, we could have used any number of intermediate frames and found that the
addition property holds for an arbitrary number of angular velocities.
We use the angular-velocity addition property a great deal in later chapters. It
should be easy to remember, because it matches physical intuition. For example,
consider a rotating disk on top of which is a second disk rotating with respect to
the first; the top disk’s rotation rate relative to absolute space is the sum of its rotation
rate and the rotation rate of the first disk.
Finally, when the various frames in a problem are attached to rigid links, as in
Section 8.1, the relative velocities of the origins are zero, and we have the multipleframe generalization of Eq. (8.5):
I
vP/O = IωA × rO /O + IωB × rO /O + IωC × rP/O .
Example 8.7 A Three-Bar Linkage
Figure 8.10 shows three bars attached to one another and to two fixed points. The
bars are allowed to pivot smoothly about their attachment points. We define three
body frames: A, which rotates with link AB; B, which rotates with link BC; and
C, which rotates with link CD. All the frames are right handed, so the out-of-plane
component is always in the direction of e3 = e1 × e2. The links move only in the plane,
so all angular velocities are in the e3 direction.
Suppose we know the initial orientation of all the links and the angular velocity
IωA of link AB. We can then solve for the angular velocity AωB of link BC relative
to link AB and for BωC of link CD relative to link BC. Since only one angular
velocity is given, we need a constraint that will allow us to find the other two.
This constraint comes from the fixed pivots A and D—namely, that the inertial
A
D
A
I
B
e2
e1
B
C
C
Figure 8.10 Three-bar linkage.
310
CHAPTER EIGHT
velocity IvD/A must equal zero. Noting that rD/A = rD/C + rC/B + rB/A, we have
Iv
I
I
I
D/A = vD/C + vC/B + vB/A = 0. We use the body frames to write
I
vB/A = AvB/A + IωA × rB/A
vC/B = BvC/B + IωB × rC/B
I
vD/C = CvD/C + IωC × rD/C .
I
Since these are rigid links, all body-frame velocities are zero:
vD/A = IωA × rB/A + IωB × rC/B + IωC × rD/C = 0.
I
(8.20)
From the angular-velocity addition property, we know that
I B
ω = IωA + AωB
I C
ω = IωA + AωB + BωC .
Since the cross product is distributive over addition, Eq. (8.20) can be written as
0 = IωA × rB/A + rC/B + rD/C + AωB × rC/B + rD/C + BωC × rD/C
= IωA × rD/A + AωB × rD/B + BωC × rD/C .
We now have one equation with two unknowns, the angular rates AωB and BωC .
Fortunately, this is a vector equation, which means we can decompose it into multiple
scalar equations. We do so by dotting with e1 and e2. Since the dot product is
distributive over addition, we need only look at one of these terms, say, the first one.
Using the scalar triple product identity (see Appendix B.4.2) yields
e1 .
ω × rD/A = rD/A . e1 × IωA .
I A
Recall that due to our frame definitions, all angular velocities are along the e3 =
e1 × e2 direction, so
rD/A . e1 × IωA = −IωA rD/A . e2 .
Applying this procedure to all terms for both the e1 and e2 directions produces two
equations:
0 = IωA rD/A . e2 + AωB rD/B . e2 + BωC rD/C . e2
0 = IωA rD/A . e1 + AωB rD/B . e1 + BωC rD/C . e1 .
RELATIVE MOTION IN A ROTATING FRAME
311
We now have two separate equations. Given the lengths and orientations of the
three links, we can easily find AωB and BωC , as well as the angular velocities in the
inertial frame and the velocities of all points in the linkage.
8.3 Planar Kinetics in a Rotating Frame
This section completes our discussion of planar relative motion by examining how
to use Newton’s second law to find the equations of motion of a particle relative
to a translating and rotating reference frame. We treated a special case of this in
Section 3.6, where we examined only a frame in translation. Now that we have the
transport equation and an understanding of the kinematics of a particle relative to
a rotating frame, we can complete the picture by including rotation. Recall that the
procedure was to find the inertial acceleration in terms of the relative acceleration
and the motion of the frame and then to rewrite Newton’s second law in terms of the
acceleration relative to the moving frame. We follow a similar approach here.
We begin with Figure 8.11 and the transport equation for the inertial velocity of
particle P in terms of its velocity in frame B and the motion of B from Eq. (8.14):
I
vP/O = IvO /O + BvP/O + IωB × rP/O .
What is the inertial acceleration of P ? With the transport equation, finding the
acceleration of P in I in terms of its acceleration in B and the angular velocity of B
in I is easy. We just apply the transport equation again to the velocity IvP/O :
I
aP/O =
Id
dt
I
vO /O +
= IaO /O +
Bd
Id
dt
vP/O + IωB × rP/O B
B
vP/O + IωB × rP/O dt
+ IωB × BvP/O + IωB × rP/O .
FP
I
B
e2
P
b2
rP/O′
rP/O
b1
O′
O
e1
Figure 8.11 Particle P subject to force FP moving relative to frame B, which may rotate or
translate relative to frame I .
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CHAPTER EIGHT
By distributing through the quantities in parentheses and using the product rule, we
find the expression for the acceleration:
I
aP/O = IaO /O + BaP/O +
+ IωB ×
The quantity
Bd
Bd
dt
I B
ω
× rP/O + 2IωB × BvP/O I B
ω × rP/O .
IωB
is the angular acceleration and, for this planar case, is equal to
Id I B Bd I B
θ̈e3. The symbol IαB =
= dt ω is often used for the angular acceleration.
dt ω
This gives us our final expression for the inertial acceleration of P :
dt
aP/O = IaO /O + BaP/O + IαB × rP/O + 2IωB × BvP/O + IωB × IωB × rP/O .
I
(8.21)
Eq. (8.21) is a general expression for the inertial acceleration of a point P in terms
of its velocity and acceleration in a translating and rotating frame. It is this expression
that we use when applying Newton’s second law. Just as in the discussion of the
polar frame in Section 3.5, the extra terms are purely kinematic. They arise because
we are trying to describe motion relative to a noninertial frame. The fourth term,
2IωB × BvP/O , is the Coriolis acceleration and the last term, IωB × IωB × rP/O , is
the centripetal acceleration. The polar frame equations in Section 3.5 are just special
cases of Eq. (8.21). The difference here is that frame B can rotate at an arbitrary
angular velocity and point P can move in B.
As in Section 3.6, we can find a form of Newton’s second law that allows us to
write the equations of motion for the coordinates of P in frame B. We just substitute
from Eq. (8.21) into Newton’s second law and solve for BaP/O , which yields
mP BaP/O = FP − mP IaO /O − mP IαB × rP/O − 2mP IωB × BvP/O (8.22)
− mP IωB × IωB × rP/O .
This equation is similar to Eq. (3.57) and is, in fact, a more general version of it. In
the absence of rotation of the body frame B, Eq. (8.22) reduces exactly to Eq. (3.57).
This equation tells us how to formulate the equations of motion for a particle relative
to a translating and rotating frame. As before, the kinematic terms appear as fictional
forces on the right-hand side. We don’t include fictional forces on a free-body diagram;
rather, we encourage using the kinematic expression for acceleration in Eq. (8.21) in
Newton’s second law.
Now we revisit the discussion in Section 3.5 and explore the physics a bit more. For
example, we can find the force required to keep a particle P stationary in a translating
RELATIVE MOTION IN A ROTATING FRAME
I
313
e2
b1
b2
B
I
ωB = Ω = θ
I
vP/O
P
O
e1
Figure 8.12 Point P traveling at a constant inertial velocity.
and rotating frame B. Suppose, for instance, you are on a spinning platform at a
playground. What force is required to stand on the platform? Since B is rotating,
P is accelerating in I and thus a force is required. If B is neither translating nor
rotationally accelerating, then IaO/O = 0 and IαB = 0. Since P is fixed in B, we
have BvP/O = BaP/O = 0, and the Coriolis term is zero. Using Eq. (8.22), the force
required to keep P fixed in B is
FP = mP IωB ×
I B
ω × rP/O .
This new, more general, expression for the kinematics in Eq. (8.22) also lets us
more easily look at a slightly different scenario that lends additional insight into the
geometric origins of the Coriolis and centripetal terms. Consider a particle P moving
at a constant velocity in the e1 direction of the inertial frame, as shown in Figure 8.12.
A body frame B with origin O = O is rotating at a constant rate with angular
velocity IωB = e3. Just as in Figure 3.23, the particle, though moving at constant
inertial velocity, is accelerating relative to frame B.
Since there are no forces acting on the particle, we can find the acceleration relative
to the body frame from Eq. (8.22):
aP/O = −2IωB × BvP/O − IωB ×
B
I B
ω × rP/O ,
where we have used the assumption that the angular velocity is constant. Again, the
particle has zero inertial acceleration because there are no forces acting on it. But due
to the kinematics of a rotating frame, P is accelerating relative to the body frame. We
explore this idea more deeply in Example 8.8.
Finally, we can double-check that Eqs. (8.14) and (8.21) lead to the right answer for
the polar frame. Let B be a polar frame with origin O. The angular velocity of B with
respect to I is IωB = θ̇e3 and the angular acceleration is IαB = θ̈e3. The position of
P in B, written in terms of components in B, is rP/O = rer . Plugging these equations
into Eq. (8.14), the inertial velocity is
I
vP/O = ṙer + r θ̇eθ ,
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CHAPTER EIGHT
P
I
FP
e2
b1
b2
B
ωr
(a)
L
e1
(b)
Figure 8.13 A simple planar model for a hurricane. (a) Hurricane. (b) Reference frames.
Image (a) courtesy of Shutterstock.
where we have used BvP/O = ṙer and IωB × rP/O = r θ̇eθ . The acceleration is found
from Eq. (8.21) to be
I
aP/O = r̈er + r θ̈ eθ + 2ṙ θ̇eθ − r θ̇ 2er
= (r̈ − r θ̇ 2)er + (2ṙ θ̇ + r θ̈)eθ .
These are the same results as Eqs. (3.26) and (3.27).
Example 8.8 Hurricanes in Flatland1
This example explores a simple model of a hurricane by restricting the phenomenon to
a planar earth. The dynamics of a real hurricane are substantially more complicated
because the earth is a rotating sphere and the physics are more sophisticated than
we treat here. However, a simple planar model is an excellent representation of the
phenomenology of hurricanes and how the Coriolis effect acts to produce the classic
spiral pattern.
A hurricane, as shown in Figure 8.13a, forms when air in the upper atmosphere
is pushed toward a low-pressure system. Figure 8.13b shows our simple model of a
hurricane. The origin of the inertial frame I = (L, e1, e2, e3) is at the center of the
low pressure. The body frame B = (L, b1, b2, b3) is our planar model of the earth’s
surface and is rotating at angular velocity IωB = ωr e3 with respect to an inertial frame.
We assume that the upper-atmosphere air mass is not rotating with the earth.
Using a multiparticle model for the air, each air particle P is attracted to the center
of the low pressure (or pushed by the surrounding high-pressure air mass) by a central
force with magnitude fL:
FP = −fLr̂P/L,
1 Flatland:
A Romance in Many Dimensions is a satirical novel by Edwin Abbott Abbott written in 1884.
Originally written as social commentary, it has become popular among scientists and mathematicians for
its fanciful treatment of dimension.
RELATIVE MOTION IN A ROTATING FRAME
315
where r̂P/L is a unit vector pointing from L to P . The particle thus accelerates toward
L by Newton’s second law:
f
I
aP/L = − L r̂P/L.
mP
What is the equation of motion for particle P in frame B? We use Eq. (8.22) to rewrite
the equation of motion in terms of the acceleration in the earth-fixed frame:
B
aP/L = −
fL
r̂P/L − 2ωr b3 × IvP/L − ωr2b3 × (b3 × rP/L).
mP
To gain insight into the motion of the particle in B, we use polar coordinates (r1, θ)B
and polar frame C = (L, er , eθ , e3). The resulting equation of motion is
(r̈ − r θ̇ 2)er + (2ṙ θ̇ + r θ̈)eθ = −
fL
er − 2ωr ṙeθ + 2ωr r θ̇er + ωr2rer .
mP
This vector equation corresponds to the following two scalar equations of motion:
r̈ − r θ̇ 2 − 2ωr r θ̇ − ωr2r = −
fL
mP
ṙ
ṙ
θ̈ + 2 θ̇ + 2ωr = 0.
r
r
A solution to these equations exists for θ̈ = 0. Solving the second equation for θ̇
yields θ̇ = −ωr . Substituting this result into the first equation leaves the equation of
motion
f
r̈ = − L .
mP
Thus each air particle rotates about the center of low pressure at a rate equal and
opposite to the earth’s rotation rate while accelerating inward toward the center. This
motion forms a spiral pattern in the earth-fixed frame.
Example 8.9 The Simple Double Pendulum
This example uses the expressions developed for kinematics in a rotating frame to
solve for the equations of motion of the simple double pendulum shown in Figure 8.14.2 A simple double pendulum has two particles: the first, labeled P in
Figure 8.14a, is suspended from a massless rod, as in the simple pendulum; the
second particle, labeled Q in Figure 8.14a, is hanging from a massless rod attached
to P . The double pendulum is a fascinating dynamical system capable of producing
chaotic trajectories, which exhibit sensitive dependence on the initial conditions.
The simple double pendulum is a two-degree-of-freedom system. We choose as
coordinates the two angles θ1 and θ2 shown in Figure 8.14a. These are the angles each
rod makes with the vertical. The angle that the lower rod makes with the first rod
2 A nonsimple double pendulum, or just the double pendulum, consists of two rigid rods hinged together.
You are invited to solve the double pendulum in Problem 9.9.
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CHAPTER EIGHT
b2 B
O
I
O
e2
g
b1
l1
e1
θ1
c2
P
P
P
c1
l2
θ2
–F1b1
C
Q
mPge1
F2c1
–F2c1
Q
Q
β = θ 2 – θ1
mQge1
(a) Coordinates
(c) Free-body diagrams
(b) Reference frames
Figure 8.14 Solving the simple double pendulum.
is labeled β = θ2 − θ1 in Figure 8.14a. Our approach to solving this problem is very
similar to the one taken for the simple pendulum in Example 3.8. Here we use the three
reference frames shown in Figure 8.14b: the inertial frame I = (O, e1, e2, e3) and
two rotating frames B = (O, b1, b2, b3) and C = (P , c1, c2, c3), where e3 = b3 = c3.
Using the coordinates shown in Figure 8.14a, the transformation tables are
b1
b2
e1
e2
c1
c2
cos θ1
− sin θ1
sin θ1
cos θ1
cos β
sin β
− sin β
cos β .
(8.23)
We use the angular-velocity addition property to find the inertial angular velocity
of each rotating frame:
I B
ω = θ̇1b3
(8.24)
I C
ω = IωB + BωC
= θ̇1b3 + β̇c3 = θ̇2c3.
(8.25)
The kinematics of point P are exactly the same as in Example 3.11,
rP/O = l1b1
I
vP/O = l1θ̇1b2
I
aP/O = l1θ̈1b2 − l1θ̇12b1.
(8.26)
These equations are found by taking the derivative of the polar-frame unit vectors
using the angular velocity from Eq. (8.24).
To find the acceleration of mass Q, we could follow the same procedure, using the
position rQ/O = l1b1 + l2c1. Instead we use Eq. (8.21), the general expression for the
acceleration in rotating frames:
RELATIVE MOTION IN A ROTATING FRAME
317
I
aQ/O = IaP/O + CaQ/P + IαC × rQ/P + 2IωC × CvQ/P
+ IωC × IωC × rQ/P .
Using the angular velocity from Eq. (8.25), the acceleration of P from Eq. (8.26),
and the fact that CaQ/P = CvQ/P = 0, we have
I
aQ/O = l1θ̈1b2 − l1θ̇12b1 + l2θ̈2c2 − l2θ̇22c1.
To find the equations of motion, we need the accelerations and forces expressed as
components in a single frame. In this problem it is easiest to choose frame B, making
the inertial acceleration of Q equal to
I
aQ/O = −(l1θ̇12 + l2θ̈2 sin β + l2θ̇22 cos β)b1 + (l1θ̈1 + l2θ̈2 cos β − l2θ̇22 sin β)b2.
Now we use Newton’s second law for each mass and the free-body diagrams in
Figure 8.14c:
−F1b1 + F2c1 + mP ge1 = mP (l1θ̈1b2 − l1θ̇12b1)
−F2c1 + mQge1 = mQ(−(l1θ̇12 + l2θ̈2 sin β + l2θ̇22 cos β)b1
+ (l1θ̈1 + l2θ̈2 cos β − l2θ̇22 sin β)b2).
Using the transformation tables in Eq. (8.23), we turn these two vector equations into
the following four scalar equations in the four unknowns F1, F2, θ̈1, and θ̈2:
−F1 + F2 cos β + mP g cos θ1 = −mP l1θ̇12
F2 sin β − mP g sin θ1 = mP l1θ̈1
−F2 cos β + mQg cos θ1 = −mQ(l1θ̇12 + l2θ̈2 sin β + l2θ̇22 cos β)
−F2 sin β − mQg sin θ1 = mQ(l1θ̈1 + l2θ̈2 cos β − l2θ̇22 sin β).
With some algebra, these equations can be solved for the equations of motion of the
two degrees of freedom:
g μQ cos β sin θ2 − sin θ1 + μQl1 sin β cos β θ̇12 + μQl2 sin β θ̇22
θ̈1 =
l1 μP + μQ sin2 β
θ̈2 = −
l1 sin β θ̇12 + l2μQ sin β cos β θ̇22 + g cos θ1 sin β
,
l2 μP + μQ sin2 β
where μP = mP /(mQ + mP ) and μQ = mQ/(mQ + mP ). Problem 8.15 invites you
to explore numerical solutions to these equations.
318
CHAPTER EIGHT
8.4 Tutorials
Tutorial 8.1 The Watt Linkage
A common dynamics problem is to convert rotation about an axis into the linear
translation of a point. For example, a rotating motor may be used to translate the
cylinder piston in an engine; another example is how a steering wheel is used to turn
the wheels of a car. The first such device was invented and patented by James Watt 3 in
1784 and used linkages. It has come to be known as the Watt linkage. Watt’s original
design is shown in Figure 8.15. A simplified model of the Watt linkage is shown in
Figure 8.16a. By rotating link A up and down about point A, the point P at the center
of link B performs (almost) straight-line motion up and down for a large range of
angles. We validate this claim using the results of this chapter.
We begin by defining three body frames: frame A, which rotates with link A
connecting points A and B; frame B, which rotates with link B connecting points B
and C; and frame C, which rotates with link C connecting points C and D. The outof-plane components of each frame are parallel to e3. The endpoints of the linkage
are fixed in the inertial frame, so we have the constraints
I
vA/O = 0
I
vD/A = 0.
(8.27)
(8.28)
Using the vector triad rD/A = rB/A + rC/B + rD/C , we can write the velocity of D
with respect to A in the inertial frame as
I
vD/A = IvB/A + IvC/B + IvD/C .
Each of these intermediate velocities can be rewritten using the transport equation:
I
vB/A = AvB/A + IωA × rB/A
vC/B = BvC/B + IωB × rC/B
I
vD/C = CvD/C + IωC × rD/C ,
I
where IωA = ωAe3, IωB = ωB e3, and IωC = ωC e3.
Assuming the links are rigid, all body velocities are zero. Using Eq. (8.28), this
assumption gives
vD/A = 0 = IωA × rB/A + IωB × rC/B + IωC × rD/C
I
= ωAl1a2 + ωB l2b1 + ωC l3c2
= θ̇AB l1a2 + θ̇BC l2b1 + θ̇CD l3c2,
3 James Watt (1736–1819) was a Scottish engineer known for his many innovative mechanical devices,
including the Watt flyball governor and the Watt linkage.
RELATIVE MOTION IN A ROTATING FRAME
319
(a)
(b)
Figure 8.15 (a) The original Watt linkage for parallel motion. (Reprinted with the permission
of Cambridge University Press.) (b) Its implementation in the Ecton mine engine.
where θAB , θBC , and θCD represent the angle each frame makes with respect to the
inertial frame (assuming right-handed rotations). From this vector equation, we find
two scalar equations by expressing all unit vectors as components in the inertial frame:
−θ̇AB l1 sin θAB + θ̇BC l2 cos θBC − θ̇CD l3 sin θCD = 0
θ̇AB l1 cos θAB + θ̇BC l2 sin θBC + θ̇CD l3 cos θCD = 0.
ωA
B
A
Ꮽ
Ꮾ
P
D
Ꮿ
C
ωC
(a)
A
I
B
b2
b1
a2
θBC
a1
e2
θAB
l1
l2
C
c2
e1
l3
O
–θCD
c1
(b)
Figure 8.16 The Watt linkage converts rotational motion into translational motion. (a) Wattlinkage schematic. (b) Reference frames.
320
CHAPTER EIGHT
1.0
0.5
meters
0.0
–0.5
–1.0
–1.5
–2.0
0
1
2
3
meters
4
5
6
Figure 8.17 The trajectory of the center of the middle link of a Watt linkage as the left link is
rotated between ±30◦.
Now assuming we know the trajectory (or the constant angular velocity) of any
of the three frames, we can integrate to find the others. Assuming that we control
θ̇AB = ωA, the equations of motion become
l1 sin θAB − θCD
θ̇BC = ωA
l2 cos θBC − θCD
l1 cos θAB − θBC
.
θ̇CD = −ωA
l3 cos θBC − θCD
(8.29)
(8.30)
To integrate Eqs. (8.29) and (8.30), we must select a trajectory for ωA that avoids
singularities in these equations. A singularity occurs when:
θBC − θCD =
π
.
2
Physically this condition corresponds to when B is parallel to either one of the other
two links. Figure 8.17 shows a trajectory of point P , the center of link B, as it is
rotated between roughly π/2 and −π/2. For this solution, link A is rotated at a rate
of ±0.5 rad/s between ±30◦ and links A , B, and C have lengths of 3 m, 1 m, and
3 m, respectively.
Tutorial 8.2 Placing a Glass on a Table Using a Three-Link Robot Arm
This tutorial revisits the three-link robot arm introduced in Example 2.5. As described
there, the arm is maneuvered by controlling the angle (and thus the angular rate) of
each joint relative to the previous link. Our goal is to determine the angular velocities
of each joint needed to place a glass on a table. The glass should follow a linear
RELATIVE MOTION IN A ROTATING FRAME
321
l1
h
l2
45°
d1
d 2 l3
(a)
b2
O′
a2
I
A
O
B
θ2
C
e2
a1 b
1
θ1 e1
O″
I
P
vP/O
(b)
Figure 8.18 A three-link robot arm places a glass of water, labeled P , on a table at constant
velocity. (a) Initial (white) and final (gray) configurations. (b) Reference frames.
trajectory at a constant velocity, as shown in Figure 8.18. We also require that the
“wrist” link be level with the ground so that no water spills from the glass.
The geometry of the problem is shown in Figure 8.18a. Because the glass must
travel an equal height and horizontal distance, we can write the components of its
velocity in the inertial frame as
√
2
2
u 0 e2 ,
=
u0 e1 +
2
2
√
I
vP/O
(8.31)
where u0 is the speed of the glass. We can use Eq. (8.18) to find IvP/O again, this time
in terms of the angular velocity of each arm relative to the inertial frame:
I
vP/O = IωA × rO /O + IωB × rO /O + IωC × rP/O ,
=0
where the body velocities are zero and the last term is zero because of the constraint
that the glass remains level. From the geometry in Figure 8.18, we have
rO /O = l1a1 =
d12 + h2a1
rO /O = l2b1 =
d22 + h2b1.
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CHAPTER EIGHT
Because all motion in this system is planar, IωA = IωA a3, and IωB = IωB b3 so that
the velocity of P is
vP/O = l1IωA a2 + l2IωB b2.
I
(8.32)
If we define θ1 as the angle of A with respect to I and θ2 as the angle of B with respect
to I as shown in Figure 8.18, the transformation arrays between A and I and B and
I are, respectively,
a1
a2
e1
e2
cos θ1
− sin θ1
sin θ1
cos θ1
and
b1
b2
e1
e2
cos θ2
sin θ2
− sin θ2
cos θ2 .
Note, θ̇1 = IωA and θ̇2 = −IωB . At the initial time t0 shown in Figure 8.18a, the
transformation arrays are
a1
a2
e1
e2
d1/ l1
−h/ l1
h/ l1
d1/ l1
and
b1
b2
e1
e2
d2 / l2
h/ l2
−h/ l2
d2 / l2 ,
and the velocity in Eq. (8.32) expressed as components in the inertial frame is
I
vP/O (t0) = h −IωA (t0) + IωB (t0) e1 + d1IωA (t0) + d2IωB (t0) e2. (8.33)
Comparing Eq. (8.33) to Eq. (8.31) gives two equations for the unknown inertial
angular velocities at t0:
√
2
I B
I A
u0
h − ω (t0) + ω (t0) =
2
√
2
I A
I B
u0 ,
d1 ω (t0) + d2 ω (t0) =
2
whose solutions are
√
2 1 − d2 / h
I A
ω (t0) =
(8.34)
u0
d1 + d 2
2
√
2 u0 I A
I B
ω (t0) =
+ ω (t0).
(8.35)
2 h
Note that these angular velocities will only give a 45◦ trajectory at the very
beginning of this system’s motion (because d1, d2, and h change as the arms move).
To find the angular velocities of the robot’s links for t > t0, we use the constraint that
the inertial acceleration of the glass is zero. Returning to Eq. (8.32), we differentiate
again to find
I
aP/O = l1 Iα A a2 − (IωA )2a1 + l2 Iα B b2 − (IωB )2b1 ,
where Iα A and Iα B are the angular accelerations. Expressing the unit vectors as
components in the inertial frame and using IaP/O = 0 produces
RELATIVE MOTION IN A ROTATING FRAME
323
−l1Iα A sin θ1 − l1(IωA )2 cos θ1 + l2Iα B sin θ2 − l2(IωB )2 cos θ2 = 0
l1Iα A cos θ1 − l1(IωA )2 sin θ1 + l2Iα B cos θ2 + l2(IωB )2 sin θ2 = 0.
With some algebra, we can solve for the two angular accelerations:
I A
−l1(IωA )2 cos β − l2(IωB )2
,
l1 sin β
I B
l1(IωA )2 + l2(IωB )2 cos β
,
l2 sin β
α =
α =
where β = θ1 + θ2.
We now numerically integrate these first-order differential equations to find
IωA (t), IωB (t), θ (t), and θ (t). The initial conditions for this integration are based
1
2
on the initial geometry of the system and Eqs. (8.34) and (8.35).
Assuming that d1 = 0.5 m, d2 = 0.12 m, and h = 1m, suppose we want the velocity
of the glass (u0) to be 1 m/s. For these values, the initial angular velocities are
IωA (t ) = 1.00 rad/s and IωB (t ) = 1.71 rad/s. The resulting trajectories of the link
0
0
angles and angular velocities are shown in Figure 8.19.
Finally, we can use the angular-velocity addition property to find the angular
velocity of each arm relative to the previous arm:
ω = IωA + AωB
I B
I C
ω = IωA + AωB + BωC = 0.
Those two equations imply
ω = −IωA − AωB .
B C
1.6
0.5
θ1
θ2
1.4
0.3
θ (rad/s)
1.2
θ (rad)
θ1
θ2
0.4
1.0
0.8
0.2
0.1
0.0
0.6
0.4
–0.1
0.2
–0.2
0
1
2
3
t (s)
(a)
4
5
0
1
2
3
4
5
t (s)
(b)
Figure 8.19 Trajectories of the angles and angular velocities of a three-link robot arm. The
initial angular velocities have been scaled so that the entire trajectory takes exactly 5 s to
execute. (a) Angles. (b) Angular velocities.
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CHAPTER EIGHT
Tutorial 8.3 The Circular Restricted Three-Body Problem
Tutorial 7.1 outlined the solution to the two-body problem in orbital mechanics and
showed that even when both bodies are free to move in absolute space, the solution
for the motion of one body relative to the other is a conic section. The center of mass
of the two bodies is either fixed or moves at constant inertial velocity.
Unfortunately, the more general problem of many bodies moving under their
mutual gravitational attraction (the N -body problem) is not solvable: there is no
simple way to write the trajectory of each body in terms of elementary functions.
Even for just three bodies, the motion is extremely complex and unpredictable,
except for some very specific arrangements (known as choreographies). In fact, it was
Poincaré’s4 study of the three-body problem that led to the development of dynamical
systems theory. Although a detailed discussion of the three-body problem and chaos
is outside our scope, a simple analysis—called the circular restricted three-body
problem—is an excellent example of the use of rotating frames for writing equations
of motion.
The circular restricted three-body problem refers to the problem in which two large
bodies (as in Tutorial 7.1) are in a circular orbit about each other and we wish to find
the approximate equations of motion for a small third body of mass m3 m1, m2.
The third body might be a spacecraft or an asteroid. In fact, several space missions
have placed a satellite into an orbit in the earth/moon or sun/earth system where the
gravitational attraction of both large bodies is important. Since we assume the third
body has negligible mass, it does not affect the circular motion of the other two bodies
(hence the adjective “restricted”).
The situation is depicted in Figure 8.20. Since there are no external forces acting
on bodies 1 and 2, we can consider the motion of body 3 relative to an inertial frame
I = (G, e1, e2, e3) fixed to the center of mass G of bodies 1 and 2. We attach a rotating
frame B = (G, b1, b2, b3) to G with b1 along the line connecting m1 and m2, and b3
parallel to e3 (and perpendicular to the plane of motion), so that b2 = b3 × b1, as
shown in Figure 8.20b. Because there are no external forces and body 1 and body 2
are in a circular orbit with fixed diameter, the total angular momentum of body 1 and
body 2 about G is conserved and frame B rotates at a constant rate relative to I with
angular velocity IωB = ne3. (The common shorthand for the rotation rate of a circular
orbit is the variable n.)
The free-body diagram for mass m3 is shown in Figure 8.20c; the only two forces
acting on mass m3 are the gravitational attractions of masses m1 and m2. Also note
that the center of mass G is not accelerating. This lets us write Newton’s second law
for the small third mass as
m3Ia3/G = −
Gm1m3
Gm2m3
r3/1 −
r3/2.
3
r3/1
r3/23
(8.36)
We are interested, however, in the motion of m3 relative to reference frame B
rotating with the two larger bodies. We therefore replace the acceleration on the lefthand side of Eq. (8.36) with the expression in Eq. (8.21):
4 Jules-Henri Poincaré (1854–1912) was a French mathematician, physicist, and philosopher. He made
seminal contributions in topology, celestial mechanics, relativity, electromagnetism, and many other areas
of mathematics and physics. He is considered the father of modern dynamical systems theory.
RELATIVE MOTION IN A ROTATING FRAME
325
m2
r3/2
I
e2
m3
I
r3/G
G
r2/O
r1/2
B
b2
m2
r3/1
e2
r3/2
rG/O
m1
r1/O
O
e3
m3
r3/G
I
ωB
e1
e1
r3/1
m1
(a)
(b)
Gm2m3
r3/2
–—
储r3/2 储 3
b1
m3
Gm1m3
r3/1
–—
储r3/1储 3
(c)
Figure 8.20 (a) Two bodies of masses m1 and m2 in a circular orbit about their common center
of mass G in the plane orthogonal to e3. A third, small body of mass m3 m1, m2 orbits both
of them. (b) Reference frames. (c) Free-body diagram for m3.
a3/G + 2IωB × Bv3/G + IωB ×
B
I B
ω × r3/G = −
μ1
μ2
r3/1 −
r3/2,
3
r3/1
r3/23
where we have substituted μ1 = Gm1 and μ2 = Gm2 and have divided by the mass m3.
At this point it is easiest to write the equations of motion in terms of specific
coordinates. We choose Cartesian coordinates in frame B, (x, y)B , so that r3/G =
xb1 + yb2. Note that the position of body 1 is r1/G = x1b1 and that of body 2 is
r2/G = −x2b1. Only motion in the plane of the two-body circular orbit is considered
here, so we ignore the z coordinate. Using IωB = ne3, Ba3/G = ẍb1 + ÿb2, r3/1 = (x −
x1)b1 + yb2, and r3/2 = (x + x2)b1 + yb2, the equations of motion in the rotating
frame are
μ1(x − x1)
μ2(x + x2)
ẍ − 2nẏ − n2x = − 3 − 3
2
2
(x + x2)2 + y 2 2
(x − x1) + y 2
(8.37)
μ1y
μ2 y
ÿ + 2nẋ − n2y = − 3 − 3 .
(x − x1)2 + y 2 2
(x + x2)2 + y 2 2
(8.38)
These equations have no general solution, and it is beyond our scope to study in
detail various possible trajectories, either analytically or numerically, though very
interesting and complex motions are possible. The important point is that we used
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CHAPTER EIGHT
the relative-motion results from Eq. (8.21) to develop equations of motion for the
coordinates of a particle in a rotating frame of reference.
There are two calculations that lend great insight into the circular restricted threebody problem without solving for exact trajectories. First, we can find an integral of
the motion in the same way we found the energy integral for many other problems.
Multiplying Eq. (8.37) by 2ẋ and Eq. (8.38) by 2ẏ and then adding the two equations
results in
2(ẍ ẋ + ÿ ẏ − n2(x ẋ + y ẏ)) = −
−
2μ1((x − x1)ẋ + y ẏ)
r13
2μ2((x + x2)ẋ + y ẏ)
,
r23
(8.39)
where r1 =
r3/1 = (x − x1)2 + y 2 and r2 =
r3/2 = (x + x2)2 + y 2. Note that
2
2
differentiating r1 and r2 yields r1ṙ1 = 2(x − x1)ẋ + y ẏ and r2ṙ2 = (x + x2)ẋ + y ẏ,
respectively. Consequently, we can multiply both sides of Eq. (8.39) by dt and
integrate to get
2
ẋd ẋ + 2
ẏd ẏ − 2n2
xdx − 2n2
ydy = −2μ1
dr1
− 2μ2
r12
dr2
r22
which equals
ẋ 2 + ẏ 2 − n2(x 2 + y 2) =
2μ1 2μ2
− C,
+
r1
r2
(8.40)
where C is a constant of integration.
Eq. (8.40) is known as the Jacobi integral, and C is Jacobi’s constant. C serves
the same purpose as the total energy, since all trajectories of body 3 must satisfy
Eq. (8.40). It is common practice to draw contours of constant C. These zero-velocity
curves are used just as the ones in Tutorial 5.3. They determine the regions, for a given
value of C, where trajectories must lie. For example, Figure 8.21 shows zero-velocity
curves for the earth-moon system over a range of values of C from 2 to 18 m2/s2. A
trajectory must lie entirely inside the curve determined by its initial condition. It is
interesting that, for initial conditions close to either large mass, the constant-potential
lines look like those for a single gravitating body and the orbits will be like those
discussed for the two-body problem in Example 4.9 and Tutorial 4.2.
The second insightful calculation is to search for equilibrium solutions of
Eqs. (8.37) and (8.38). These are points in frame B where body 3 remains stationary.
We find these points by setting the speeds and accelerations equal to zero in Eqs. (8.37)
and (8.38). There are five such points in this rotating system. Three of them lie on the
line connecting masses m1 and m2 (y = 0). Eq. (8.38) is satisfied trivially there, and
Eq. (8.37) reduces to the quintic polynomial,
n2x(x − x1)2(x + x2)2 − μ1(x + x2)2 − μ2(x − x1)2 = 0.
The three real roots of this polynomial correspond to the three equilibrium points on
the x-axis, which we label L1, L2, and L3.
RELATIVE MOTION IN A ROTATING FRAME
327
y
x
Figure 8.21 Zero-velocity curves for the earth-moon system. The earth is at the center and
the moon to the right.
The other two equilibrium points lie on the vertices of an equilateral triangle such
that r1 = r2 = r, the distance between body 1 and body 2. To see how these equilibrium
points satisfy Eqs. (8.37) and (8.38), use the identity n2 = (μ1 + μ2)/r 3 and the
center-of-mass corollary, m1x1 = m2x2. The final two equilibrium points are then
L4 =
√ r
3
∗
r
− μ r,
2
2
√
L5 =
B
3
r
− μ∗r, −
r
2
2
,
B
where μ∗ =
m1/(m1 + m2) is the mass ratio of the large bodies.
These five points were originally found by Lagrange5 in 1772 and are now often
referred to as the Lagrange points. They are schematically shown in Figure 8.22. It
can be shown that the collinear points L1, L2, and L3 are all unstable, whereas, for
certain ratios of m1/m2, the equilateral points L4 and L5 are stable.6 For instance,
the sun/Jupiter L4 and L5 points are stable equilibrium locations in the sun/Jupiter
system. It was a great verification of the theory when asteroids were discovered to be
residing at these points more than 100 years after Lagrange’s first calculation.
5 Joseph-Louis Lagrange (1736–1813) was an Italian-French mathematician and astronomer known for his
many contributions to mechanics.
that, if a particle starts near a stable equilibrium point, then it stays near that point for all time,
whereas if a particle starts near an unstable equilibrium point, it is not guaranteed to stay nearby.
6 Recall
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CHAPTER EIGHT
L4
X
r
r
L3
L1
X
X
r
m1
L2
X
m2
r
r
X
L5
Figure 8.22 Schematic of the five Lagrange points in the circular restricted three-body
problem.
8.5 Key Ideas
.
If P is a point fixed to a rigid body B, the inertial velocity of P with respect to
O is
I
vP/O = IvO /O + IωB × rP/O ,
where B is a reference frame fixed to B at point O , and IωB is the angular velocity
of B with respect to I .
.
.
Under the no-slip condition, the point of a rolling disk or wheel in contact with
the ground has zero velocity relative to the ground.
The transport equation can be used to find the time derivative of a vector c with
respect to frame A:
Ad
dt
c=
Bd
dt
c + AωB × c.
For example, the inertial velocity of P with respect to the origin O of a rotating
frame B is
Bd rP/O =
rP/O + IωB × rP/O .
dt
dt
Id .
Angular velocities obey the addition property:
I B
ω = IωA + AωB ,
where A and B are rotating reference frames.
.
A general expression for the inertial acceleration of P with respect to O in terms
of its velocity and acceleration in a rotating frame B is
RELATIVE MOTION IN A ROTATING FRAME
329
I
aP/O = IaO /O + BaP/O + IαB × rP/O + 2IωB × BvP/O + IωB × IωB × rP/O ,
where O is the origin of B and IαB =
eration of B with respect to I .
Bd
dt (
IωB ) = Id (IωB )
dt
is the angular accel-
8.6 Notes and Further Reading
Without question, the most important result of this chapter is the transport equation,
relating, by means of the angular velocity, the derivative of a vector in one frame to the
derivative of the same vector with respect to another frame. In one notation or another,
the transport equation appears in every dynamics textbook, from introductory to
advanced. It is the cornerstone of kinematics in rotating frames. Surprisingly, given its
importance and ubiquitous application, there is no generally accepted nomenclature
for referring to it. Most textbooks don’t use a name at all (e.g., Greenwood 1988; Moon
1998; Meriam and Kraige 2001; Bedford and Fowler 2002; Hibbeler 2003; Tongue
and Sheppard 2005; Beer et al. 2007). Kane (1978) discusses what he calls kinematic
theorems, the transport equation being the first (the second refers to vector summation
of velocities). The later book Kane and Levinson (1985) drops the reference and
simply discusses “differentiation in two reference frames.” Tenenbaum (2004) also
refers to this equation as a kinematic theorem. Rao (2006) calls it the transport
theorem, which unfortunately can cause confusion with the well-known Reynolds’
transport theorem in fluid mechanics. It is very convenient to be able to reference this
equation by name, so we refer to it as the “transport equation.” We will often refer to
it by name rather than by an equation number to get you used to its importance, just
as we refer to Newton’s second law.
This chapter also presented several important engineering examples that appear in
most texts and arise frequently in practice, including linkages, gears, and wheels. We
recommend many texts for further exploration of examples and problems, including
Moon (1998), Meriam and Kraige (2001), Bedford and Fowler (2002), Hibbeler
(2003), Tongue and Sheppard (2005), Beer et al. (2007), and Ginsberg (2008). Our
particular examples are very similar to those in Bedford and Fowler (2002).
James Watt was a singularly important figure in the history of engineering. His
inventions were transformational and were instrumental in ushering in the industrial
revolution; many are still used today. We describe two in detail in this book (the Watt
linkage and the Watt flyball governor). Readers interested in more detail on Watt and
his work are directed to Dickinson (1936).
The three-body problem is a classic and important problem in orbital mechanics. It
was first studied in detail by Poincaré and resulted in the discovery of chaos theory. An
excellent historical discussion is in Diacu and Holmes (1996). An extremely thorough
(albeit advanced) treatment, including the restricted problem, is in Szebehely’s (1967)
classic book. Vallado (2001) also has a good introductory discussion to the restricted
problem. Those interested in a modern and much deeper treatment of the three- and
four-body problems and motion about equilibrium points are directed to Gómez et al.
(2001) and Koon et al. (2007).
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CHAPTER EIGHT
8.7 Problems
Consider a body frame B = (O , b1, b2, b3) translating and rotating with
respect to an inertial frame I , as shown in Figure 8.23. The instantaneous
position of particle P with respect to O is rP/O = b1 + 2b2. The velocity
of O with respect to O is IvO /O = 2ex m/s. The angular velocity of B with
respect to I is IωB = 3b3 rad/s.
8.1
a. If P is fixed in B, find the velocity IvP/O of P with respect to O in I .
b. If P is moving with respect to O in B at velocity BvP/O = 3b2, find
the velocity IvP/O of P with respect to O in I .
I
ey
B
b2
P
(1,2)B
b1
I
O′
O
vO′/O
ex
Figure 8.23 Problem 8.1.
8.2
Let B = (O , b1, b2, b3) be an arbitrary (planar) body frame with angular
velocity IωB relative to an inertial frame I = (O, e1, e2, e3), where e3 = b3.
I
Prove that dtd (bi ) = IωB × bi , for i = 1, 2, 3.
8.3
Consider the bicycle model in Figure 8.24. Suppose crank shaft A is pedaled
at IωA = 2 rpm and is connected to gear B by a chain of fixed length. The
gear is fixed to wheel C . If rA = 2rB = 0.5rC = 0.1 m, how fast does the bike
roll without slipping? If rB increases, does the bike roll faster or slower?
Ꮿ
rC
O″
Ꮾ
O′
rA
Ꮽ
I
O
Figure 8.24 Problem 8.3.
RELATIVE MOTION IN A ROTATING FRAME
8.4
331
An astronaut (white dot) is walking radially outward at v0 relative to a space
station, as shown in Figure 8.25. The space station is traveling with inertial
velocity IvO /O and rotating with constant angular rate . Find the magnitude
of the inertial velocity of the astronaut as a function of IvO /O , , and her
radial distance r from O .
v0
O′
I
vO′/O
Ω
Figure 8.25 Problem 8.4.
8.5
Consider a carousel rotating at angular velocity IωB relative to an inertial
frame fixed to the ground with origin O. There is a clown on the nonrotating
wall (into whose mouth you want to throw a brass ring) located at a vector
position rC/O relative to O, the center of the carousel. You are on a horse
fixed to the carousel (ignore its up-and-down motion) at position rH/O . What
is the velocity of the clown (rate of change of rC/H ) relative to you (i.e., in the
body frame of the carousel) in terms of IωB and rC/O ? (See Example 8.5.)
8.6
Suppose you are the gunner on a fighter jet traveling with inertial velocity
Iv
J/O . The ground has radioed to you an incoming threat with inertial velocity
Iv
T/O (assume that the earth is an inertial frame). To fire and hit the threat,
you need to know its velocity relative to your jet. Your jet is flying level, but it
is executing an evasive turn, so it is rotating with angular velocity IωB . Based
on knowledge of your position and information from the ground, you are able
to compute the position of the target rT/J relative to your jet. Write down the
equation you would use to compute the velocity of the target relative to a frame
fixed to your jet (B frame) in terms of IvT/O , IvJ/O , rT/J , and IωB .
8.7
Consider a massless disk with a slot, as shown in Figure 8.26. The slot is
offset from the diameter by distance l, and the disk is rotating with constant
angular velocity . Inside the slot is a particle P of mass m connected to a
linear spring with spring constant k. At time t = 0 the spring is stretched an
amount x0 (assume that the unstretched length of the spring is zero). Note that
the disk is in a plane and there is no gravity acting.
a. Find the equation of motion for the mass position x inside the slot.
b. For what does ẍ go to zero? What physically is happening?
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CHAPTER EIGHT
x
P
l
Ω
O
r
Figure 8.26 Problem 8.7.
c. What is the normal force the disk applies to the mass in terms of m, x,
ẋ, l, and ?
d. Is energy conserved in this problem?
e. What is the angular momentum IhP/O of the point mass relative to
origin O? Is it conserved during the motion?
8.8
A particle of mass m slides inside a circular slot of radius r cut out of a massless
disk of radius R > r, as shown in Figure 8.27. The particle is attached to a
curvilinear spring with spring constant K and unstressed angle θ0, where the
angle θ describes the position of the particle relative to the attachment point
A of the spring. Knowing that the disk rotates with constant angular velocity
e3 about an axis through its center at O and assuming no gravity, determine
the following:
m
P
θ
r
O
A
Ω
Figure 8.27 Problem 8.8.
a. the velocity and acceleration of P relative to a fixed observer (you can
pick the frame used to write the components).
b. the differential equation of motion for the particle in terms of angle θ.
RELATIVE MOTION IN A ROTATING FRAME
8.9
333
Sketch a planar model of a K-MAX helicopter delivering cargo to a moving
ship, as shown in Figure 8.28, using a point mass to represent the helicopter.
Derive the inertial kinematics of the cargo with respect to a point fixed to the
deck of the ship, assuming that the length of the cargo sling is fixed.
Figure 8.28 Problem 8.9.
8.10 Consider a disk of radius R = 1that is rotating at constant angular rate = 0.2
rad/s about an inertially fixed point O, as shown in Figure 8.29. Body frame
B = (O, b1, b2, b3) is fixed to the disk. Particle P is launched with respect to
O in B at velocity BvP/O = −v0b1, where v0 = 1 m/s. There are no external
forces acting on P after it is launched.
I
B
P
–v 0b1
O
R
Ω
Figure 8.29 Problem 8.10.
a. Using the coordinates (x, y)I , find the equations of motion of P .
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CHAPTER EIGHT
b. Integrate the equations of motion in matlab for 2 s and plot position
rP/O (t) from the perspective of an observer in I . [HINT: Assume that
B and I are aligned at the instant P is launched.]
c. Using the coordinates (x, y)B , find the equations of motion of P .
d. Integrate the equations of motion in matlab for 2 s, and plot the
position rP/O (t) from the perspective of an observer in B.
8.11 Consider the locomotive wheel sketched in Figure 8.30. Assuming the wheel
rolls without slipping and the piston moves only horizontally, derive the
kinematic relationship between the velocity of a point on the piston and the
velocity of the train.
Q
d
R O′
P
Figure 8.30 Problem 8.11.
8.12 An astronaut drops a wrench while on a space walk from her space station
(based on a true story!) (Figure 8.31). Assume that there are no forces acting
on the wrench and that the space station is translating at constant velocity
Iv and rotating with respect to an inertial frame I at constant angular rate
O /O
. Let B be a body frame attached to the space station. Use the coordinates
(x, y)B to derive the kinematics of P with respect to O, and then find the
equations of motion of x and y. If the astronaut releases the wrench from rest
in B, what is the initial velocity IvP/O ?
P
B
(x, y)B
I
O′
Ω
O
Figure 8.31 Problem 8.12.
8.13 Consider two particles, each of mass m, free to slide on rails forming a right
angle and connected by a spring of spring constant k, as shown in Figure 8.32.
The spring is unstretched when each mass
√ is a distance x0 from point O (so
the unstretched length of the spring is 2x0).
RELATIVE MOTION IN A ROTATING FRAME
335
x1
O
m
x2
k
m
Figure 8.32 Problem 8.13.
a. How many degrees of freedom are there in this problem?
b. Is the linear momentum of the center of mass conserved during any
motion of the masses?
c. Write down the equations of motion for the masses.
d. Now suppose the angle bracket rotates at constant angular velocity
e3 perpendicular to the plane (with e3 going through O). What are
the new equations of motion?
e. How many equilibrium arrangements are there for the masses while
the bracket is rotating? Explain.
8.14 Consider a bead sliding down a frictionless wire under the influence of gravity
as shown in Figure 8.33. Suppose the bead starts at rest at height y = h at x = 0.
Also assume that the shape of the wire is given by the function y = f (x).
I
ey
P
g
y = f(x)
O
ex
Figure 8.33 Problem 8.14.
a. Find an expression for the velocity IvP/O (x) of the bead as a function
of horizontal position x in terms of components in the inertial frame
directions ex and ey .
b. Now suppose the wire and bead are rotating together at angular rate about the ez axis. What is the inertial velocity of the bead now?
8.15 Recall the simple double pendulum considered in Example 8.9. We now
explore some solutions to the equations of motion developed in the text.
a. Write a matlab function that accepts as inputs l1, l2, mP , mQ, a time
span (initial and final times), and initial conditions for θ1, θ2, θ̇1, and θ̇2
336
CHAPTER EIGHT
and integrates the equations of motion of the simple double pendulum
for these values over the specified time span.
b. Integrate the equations of motion for mP = mQ = 4 kg, l1 = l2 =
1 m, and initial conditions such that the pendulum starts from rest
with both arms extended horizontally to the right of the attachment
point (i.e., θ1(0) = θ2(0) = π/2). Plot the first 10 s of the resulting
trajectories of points P and Q on one graph, using coordinates in
the D = (O, ex , ey , ez ) frame, where ex = e2 and ey = −e1 (these
would be the paths you observed the particles tracing out if you were
observing the double pendulum in action). Repeat the integration for
initial conditions with the first link as before and the second link
starting pointing up, perpendicular to the first (i.e., θ2(0) = −π).
c. Perform integrations with unequal masses and link lengths. Explain
how these differences affect the behavior of the double pendulum.
8.16 A particle P moves along a straight radial groove in a circular disk of radius
a that is pivoted about a perpendicular axis through its center O. The particle
moves relative to the disk so that
r=
a
(1 + sin ωt),
2
and the disk rotates according to
θ = θ0 sin ωt.
Find the general expression for the absolute acceleration (i.e., in inertial space)
of P in terms of a, θ0, ω, and t.
8.17 Suppose you are the navigator on a ship in choppy seas so that the ship is
rocking about its center of mass at rate ωr (i.e., the ship’s angle is given by
φ = a sin ωr t), as shown in Figure 8.34.
Near the bow, a distance d from the center of mass, you have a pendulum
clock. (You are in the early eighteenth century, and that is the only available
clock.) Thus time is kept based on a calibrated period of the swinging pendulum. (You can ignore the height of the pendulum above the deck of the ship;
it is exaggerated in the figure for clarity.)
Ship deck
θ
θ
ϕ = a sin ωr t
d
Figure 8.34 Problem 8.17.
a. Find the equation of motion for the angle of the clock’s pendulum
relative to its base (i.e., to a frame fixed to the ship).
b. What is the likelihood that the clock will keep accurate time while
at sea?
CHAPTER NINE
Dynamics of a Planar Rigid Body
As discussed in Chapter 8, a rigid body is a collection of many particles that are
constrained to be stationary relative to one another. In this chapter, we use the results
of Chapter 7 to find equations for the translational and rotational motion of a rigid
body. Our only restriction is that the motion is still confined to a plane. Despite
this restriction, we will encounter many realistic and useful problems and lay the
groundwork for the three-dimensional case studied in Part Four.
9.1 A Rigid Body Is a Multiparticle System
Chapters 6 and 7 discussed multiparticle systems in depth. In the most general,
unrestricted, three-dimensional case, a system of N particles has 3N degrees of
freedom. There is little we can do to simplify such a problem: a complete description
requires 3N equations of motion. We did, however, introduce some useful facts that
made solving many problems easier. For instance, the translational motion of the
center of mass of the collection can be treated as if it were a single particle of
mass mG equal to the total mass of the collection. We then devoted our attention
to understanding the motion of the collection relative to the center of mass. Even that
was not a great simplification, however, as it still involved the same large number of
equations. We thus restricted the discussion in Chapter 6 to either a small number of
particles (collisions, orbits) or to a continuum of particles (variable-mass systems).
Chapter 7 introduced additional tools to study the motion of the particles relative to
the center of mass (or some other fixed point), namely, angular momentum and energy.
These tools supply concise formulas for writing the equations of motion relative to the
center of mass. Nevertheless, the problem of scale did not necessarily go away, and
many particles meant many equations. We thus limited the discussion in that chapter
to systems with a small number of particles or ones with many constraints.
Now we turn to the special case of a planar rigid body. Figure 9.1 shows an example
of a rigid and nonrigid body. Put simply, a rigid body is a collection of particles with
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CHAPTER NINE
(a)
(b)
Figure 9.1 Illustration of (a) a rigid (frozen) fish and (b) a flexible (swimming) fish. Images
courtesy of Shutterstock.
enough constraints to reduce the degrees of freedom to the motion of only a single
object.
Definition 9.1 A rigid body is a collection of particles constrained to remain
motionless relative to one another.
Mathematically, this definition is represented by the constraints
d
ri/j = 0
dt
(9.1)
for all pairs i and j . Although this set of constraints is true for every rigid body, it
is more than the minimum number of constraining links among particles needed to
make a set of N particles rigid. (How many massless rods are necessary to connect a
set of N particles so that the resulting collection is rigid? While this is an interesting
question, its investigation is beyond the scope of this text.)
It is because of the many constraints holding the constituent particles fixed relative
to one another that the number of degrees of freedom reduces dramatically. Recall
that, in two dimensions, the number of degrees of freedom of N particles is 2N .
A two-dimensional rigid body composed of N particles, however, has only three
degrees of freedom: two degrees of freedom describe the position of the center of mass
(translation) and one describes the orientation of the body (rotation). This reduction
results from the 2N − 3 constraints of the form Eq. (9.1) that make the collection
rigid (see Eq. (2.12)).
In three dimensions, a general collection of N particles has 3N degrees of freedom.
A three-dimensional rigid body, however, has only six degrees of freedom: three
DYNAMICS OF A PLANAR RIGID BODY
339
B
b2
Ꮾ
I
i
e2
ri/G
ri/O
b1
G
rG/O
O
e1
e3
Figure 9.2 Planar rigid body B with attached body frame B translating and rotating with
angular velocity IωB with respect to fixed inertial frame I .
degrees of freedom describe the position of the center of mass and three describe
the orientation of the body. Thus 3N − 6 constraints of the form Eq. (9.1) are needed
to make the collection rigid. This result holds even for a continuous rigid body, by
which we mean a collection of N particles in the limit N → ∞.
Figure 9.2 shows a multiparticle system collected together as a rigid body labeled
B. Because the particles do not move relative to one another, we can unambiguously
attach a body frame to B, which we label B, and whose origin is located at the
center of mass G of the rigid body. In the remainder of this chapter we develop a
formalism for describing the translational motion of the body in the inertial frame and
the rotational motion of the rigid body relative to I , while still restricting ourselves to
only two dimensions and thus three degrees of freedom (two of translation and one of
rotation).
Before we move on, a word about models. This is a good time to review the
discussion in Chapter 2 on Newton’s laws. Recall that we emphasized that these three
laws were not meant to explain matter and inertia but rather to provide predictive
tools for describing motion. The concepts of point mass and force are abstractions—
useful models that when used correctly with Newton’s laws are remarkably effective
at describing the motion of particles. The same is true here in this discussion of rigid
bodies. The idea of constructing a rigid body as a continuous collection of point
masses is a useful model that allows us to apply Newton’s laws; its veracity is proven
by its effectiveness. There is no reductionalist theory that allows us to break down
rigid bodies into actual point masses. At the atomic scale, quantum mechanics takes
over and the Newtonian approach to forces and moments breaks down. Nevertheless,
as long as we use our models consistently and correctly apply the laws of motion, the
results are valid. The notes at the end of the chapter briefly discuss an alternative to
the particle or corpuscular model for a rigid body and provide some references for
further exploration.
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9.2 Translation of the Center of Mass—Euler’s First Law
Recall from Chapter 6 that the motion of the center of mass G of a collection of
particles can be treated as if it were a single particle located at the center of mass with
mass mG equal to the total mass of the collection. This result applies unchanged to
rigid bodies. In other words, we can extend Newton’s second law and apply it to the
center of mass of the rigid body. This has come to be known as Euler’s first law, as it
was Euler who first developed the equations of motion for a rigid body.
Law 9.1 Euler’s first law states that the total mass mG of a rigid body times the
inertial acceleration of the center of mass equals the total external force on the
rigid body:
FG = mGIaG/O .
Note that we have dropped the (ext) superscript on the force that appears in Eq. (6.16)
for multiparticle systems in this statement of Euler’s first law for a rigid body. Since
the internal forces in the rigid body will always cancel by Newton’s third law, we
simplify the notation for the translational equation of motion of the center of mass. It
is implied that FG represents the total external force acting on the body.
In many respects we have been using Euler’s first law all along. Most of the objects
for which we were finding trajectories were actually extended rigid bodies (e.g., a
satellite or planet in orbit). It is this law that justifies our treatment of the translation
of those bodies as if they were particles.
As discussed in Chapter 6, a consequence of Euler’s first law is that, if the total
external force is zero, then the linear momentum of the center of mass of a rigid body
is conserved.
Since we most often consider continuous rigid bodies, it is necessary to modify
Definition 6.2 for the center of mass of a continuous body. Instead of modeling the
rigid body as a collection of particles of mass mi and then summing over i, we assign
to each point of the rigid body a differential mass dm and position rdm/O relative to
O. We then convert the sum in Eq. (6.13) to an integral over B:
rG/O =
1
mG
B
rdm/O dm.
Assuming the rigid body has density ρ(rdm/O ), which may vary with position
inside the body, we can replace the differential mass element dm by the density times
the differential volume element dV to obtain the continuous formula for the center
of mass:
rG/O =
1
mG
B
rdm/O ρ(rdm/O )dV .
(9.2)
DYNAMICS OF A PLANAR RIGID BODY
341
B
b2
b2
hx
–
y=h–w
h
G
O
B
G
r2
w
r1
O
b1
(a)
b1
(b)
Figure 9.3 Computing the center of mass of a continuous rigid body. (a) Cartesian coordinates.
(b) Polar coordinates.
(We discuss finding differential area and volume elements in different coordinate
systems in Appendix A.) For example, in Cartesian coordinates, the density is
ρ(rdm/O ) = ρ(x, y, z), and the differential volume element is dV = dxdydz. (These
coordinates can be expressed in either the body frame or instantaneously in an inertial
frame, though the body frame is usually the simplest.)
For a continuum of particles, the center-of-mass corollary in Eq. (6.15) is
rdm/Gρ(rdm/G)dV = 0.
B
(9.3)
Example 9.1 Computing the Center of Mass of a Continuous
Rigid Body (Cartesian Coordinates)
Consider the right-triangular rigid body shown in Figure 9.3a. To use Eq. (9.2) to find
the center of mass, we start by writing down the position rdm/O = xb1 + yb2 and the
volume element (actually, area element in two dimensions) dV = dxdy. Note that
we express the position as vector components with respect to a body-fixed reference
frame B = (O, b1, b2, b3). Assuming mass m and uniform density ρ, we have
rG/O =
=
=
ρ
m
ρ
m
w
y
0
0
(xb1 + yb2)dydx
h
x
h− w
w
0
0
xdydx b1 +
ρ
m
ρhw 2
ρh2w
b1 +
b2 .
6m
6m
Using m = ρhw/2 results in
rG/O =
w
h
b 1 + b2 .
3
3
h
x
h− w
w
0
0
ydydx b2
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Example 9.2 Computing the Center of Mass of a Continuous
Rigid Body (Polar Coordinates)
We now use polar coordinates in a Cartesian frame to compute the center of mass of
the annular rigid body shown in Figure 9.3b. In polar coordinates, the position of a
small mass element is rdm/O = r cos θb1 + r sin θ b2 and the volume (area) element
is dV = rdrdθ. Assuming mass m and uniform density ρ, the center of mass integral
in Eq. (9.2) becomes
rG/O =
=
π
ρ
m
0
r2
(r cos θb1 + r sin θb2)rdrdθ
r1
2ρ 3
(r − r13)b2.
3m 2
Using m = (ρπ r22 − ρπr12)/2 = ρπ(r22 − r12)/2, we obtain
rG/O =
4(r23 − r13)
3π(r22 − r12)
b2.
Example 9.3 A Falling Rigid Body
In Example 6.5 we found the equations of motion for a collection of particles falling
in a uniform gravitational field and showed that they reduce to the single, simple
equation for an equivalent falling particle of mass mG located at the center of mass of
the collection, G. Here we repeat that example for a rigid body B. Figure 9.4a shows
a rigid body of arbitrary shape falling in a uniform gravitational field with acceleration
−gey . Thus every mass element dm of the rigid body is acted on by a force −dmgey ,
as shown in Figure 9.4b.
Using Euler’s first law and integrating over the external force on each mass element
dm yields:
mG
Id 2 dt 2
rG/O = −
B
dmgey = −mGgey .
(9.4)
Thus, as for the collection of free particles, the center-of-mass equation of motion is
rG/O = −gey ,
Id 2 dt 2
which is independent of the orientation of the body. As before, this complete decoupling is due to the fact that, for a constant force field, the integral of the force in
Eq. (9.4) is particularly simple.
If the force on each mass element instead depended on position (as for 1/r 2
gravity), then the problem would be more complicated and the equation of motion
DYNAMICS OF A PLANAR RIGID BODY
g
I
dm
dm
ey
–dmgey
rdm/G
rdm/O
G
rG/O
O
343
G
Ꮾ
ex
(a) Coordinates and reference frame
(b) External force on mass element dm
Figure 9.4 Planar rigid body falling freely in a uniform gravitational field.
for G would depend on the orientation of the rigid body. This situation is similar to
that discussed in Tutorial 6.1. For a general rigid body, the center of gravity is not the
same as the center of mass. Although the center of mass and center of gravity are
the same for every rigid body in a uniform gravitational field, that is not the case
in the more exact 1/r 2 field. It is only for very special geometries—like a sphere—
that they coincide. Such bodies are referred to as centrobaric.
9.3 Rotation about the Center of Mass—
Euler’s Second Law
In the last example of the previous section we explained that, for more complicated
force laws, the equation of motion of the center of mass could depend on the orientation of the rigid body without carefully defining the term orientation. What, in fact, do
we mean by orientation? And how might we find the rotational dynamics of a planar
rigid body? In general, orientation refers to the arrangement of the particles relative
to the center of mass in the inertial frame. Since the body is rigid, this is equivalent to
describing the orientation, or rotation, of body frame B in the inertial frame. In two
dimensions, the orientation of a rigid body in I has a single degree of freedom. We
typically use the coordinate θ to describe the rotation of a planar rigid body about a
single axis fixed in the inertial frame.
Again, for the planar, two-dimensional situation considered in this chapter, the
orientation of B in I is given entirely by the single coordinate θ and the rotational
motion is thus described by the angular velocity of B in I . For now, the angular
velocity is simple; that is, it is constrained to point in the e3 (and thus b3) direction
(IωB = IωB b3 = θ̇b3). In this section we search for the equation of motion for the
angular velocity, in terms of the coordinate θ, by utilizing the separation of angular
momentum described in Chapter 7. We begin as in that chapter with the equation of
motion for the total angular momentum of a rigid body.
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9.3.1 The Angular Momentum of a Rigid Body
Chapter 7 defined the total angular momentum of a collection of particles about an
inertially fixed point O (see Definition 7.1):
I
hO =
N
mi ri/O × Ivi/O .
i=1
This expression holds unchanged for a discrete rigid body. For a continuous body, we
extend it to integral form, just as for the definition of the center of mass:
I
hO =
B
rdm/O × Ivdm/O dm.
To keep things simple, we often use the summation notation for a finite number of
rigidly connected particles. Just remember that everything still applies if you instead
use an integral over a continuous body.
Taking the derivative of the angular momentum, as in Eq. (7.1), once again yields
Id
I
dt
hO =
N
ri/O × mi
i=1
Id
dt
I
vi/O ,
which, after substituting from Newton’s second law and performing a bit of manipulation, reduces to
Id
dt
I
hO =
N
i=1
ri/O × Fi(ext) +
N N
1 (ri/O − rj/O ) × Fi,j .
2 i=1 j =1
(9.5)
(ext)
=MO
At this point we are a bit stuck. We want eventually to obtain an expression just like
Eq. (7.5): the rate of change of the angular momentum of a rigid body is equal solely
to the external moments on it. In fact, as you will soon see, this is Euler’s second
law. However, to get there from our particle-dynamics model requires eliminating the
internal-moment term in Eq. (9.5). In Chapter 7 we did this by invoking the internalmoment assumption (Eq. (7.4)). The result was a compact expression, but one that held
only for multiparticle systems for which the internal-moment assumption applied.
It turns out that Newton’s laws are not enough to find the equations of motion for a
rigid body without making an additional axiomatic assumption. In fact, Euler, for this
reason, abandoned the particle model for rigid bodies entirely and simply postulated
new laws of motion for rigid bodies, just as Newton postulated his laws for particles.
Although this approach is legitimate, and one taken by some authors, we instead
postulate the internal-moment assumption as a property of rigid bodies. Recall the
comments at the end of Section 9.1: our approach to dynamics is to develop models
that, along with the laws of motion, effectively describe and predict motion. Part
of our particle-based model of a rigid body is that the internal-moment assumption
must hold. This assumption should not come as a surprise; in statics you may have
analyzed the internal axial and shear forces and internal moments of a rigid body,
always assuming that their sum was zero. The internal-moment assumption is part of
DYNAMICS OF A PLANAR RIGID BODY
345
the foundation of macroscopic material mechanics and will arise again in courses on
vibration and elasticity.
Under the internal-moment assumption, the second term in Eq. (9.5) is zero, and
we obtain Euler’s second law.
Law 9.2 Euler’s second law states that the rate of change of the inertial angular
momentum of a rigid body about point O in the inertial frame is equal to the
total external moment acting on the rigid body about O:
Id
dt
I
hO = M O .
(9.6)
Euler’s second law simply states that Eq. (7.5) for the change in total angular
momentum of a collection of particles applies to a rigid body. This fact is enormously
important and is the foundation of everything we do in the remainder of the book.
Note also that, as with Euler’s first law, we have eliminated the (ext) superscript on the
applied moment to simplify the notation, since, by our internal-moment assumption,
Euler’s second law applies only for external moments on the rigid body.
Also, as in Law 7.1, when the total external moment acting on the rigid body about
O is zero, the total angular momentum of the rigid body about O is conserved. So,
for instance, under a central force, the center of mass of a rigid body will orbit like a
particle.
Eq. (9.6) is certainly useful for rigid-body problems. However, of even more use is
the application of the angular momentum separation principle to rigid bodies. Look
back over Section 7.2. Everything done there applies equally well to a rigid body. Thus
we can write the total angular momentum of a rigid body as the sum of the angular
momentum of the center of mass plus the angular momentum about the center of
mass:
I
hO = IhG/O + IhG,
where
I
hG/O =
mGrG/O × IvG/O
(9.7)
is the angular momentum of the center of mass relative to O, and
I
hG =
N
mi ri/G × Ivi/G
(9.8)
i=1
is the angular momentum about the center of mass. For a continuous rigid body,
I
hG =
B
rdm/G × Ivdm/G dm.
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CHAPTER NINE
Just as in Chapter 7, we can take the derivative of each of these angular momenta.
Doing so for Eq. (9.7) results in
Id
dt
I
hG/O = rG/O × FG =
MG/O ,
(9.9)
where again we have removed the (ext) subscript for brevity. Eq. (9.9) is Euler’s first
law in angular-momentum form, analogous to that for a multiparticle system. Again,
Eq. (9.9) states that the center-of-mass motion of a rigid body can be considered
separately from its rotational motion; the center-of-mass motion satisfies Newton’s
second law, in either linear-momentum or angular-momentum form.
Taking the time derivative of the angular momentum about the center of mass in
Eq. (9.8) follows the same process. To eliminate the internal moments, we again have
to invoke the internal-moment assumption. Doing so results in Eq. (7.12) applied to
a rigid body:
Id
I
dt
hG = M G .
(9.10)
This equation is Euler’s second law for the angular momentum and moments about
the center of mass. The total moment MG on the rigid body in Eq. (9.10) is
MG =
N
ri/G × Fi(ext).
(9.11)
i=1
When a continuous field force is acting on the body, let fdm be the force per unit mass
on the body (also called the specific force). In this case:
MG =
B
(ext)
dm.
rdm/G × fdm
(9.12)
When dealing with a collection of individual particles (in this case, rigidly attached), Fi is the total force on the ith particle and ri/G is the position of that particle.
For a continuous rigid body, there are an infinite number of particles. In that case, for
the contact forces in Eq. (9.11), we interpret the moment equation slightly differently.
MG represents the net moment of some finite number of contact forces applied to the
rigid body at locations ri/G. It should not be surprising by now that when we consider
a rigid body (i.e., a system of particles), both the magnitude of the applied force and
the location of its application are important when solving for the complete motion
(translation and rotation). Likewise, the moment due to field forces, such as gravity,
can depend on shape, location, and orientation. We highlight these observations in
the next few examples.
DYNAMICS OF A PLANAR RIGID BODY
347
Example 9.4 The Moment on a Rigid Body Due to Uniform Gravity
Here we reconsider the rigid body falling in a uniform gravitational field from Example 9.3 and shown in Figure 9.4. Example 9.3 showed that we could treat the rigid body
as a single particle located at the center of mass with the force of gravity, −mGgey ,
acting on it. Here we ask whether there is a moment generated on the rigid body by
the gravitational field, thus allowing us to separately write a rotational equation of
motion. From Figure 9.4b, the force per unit mass on each mass element is just −gey .
The external moment from Eq. (9.12) is thus
MG = −
B
rdm/G × gey dm.
Since g and ey are independent of position, we can factor them out of the integral
to find
MG = −g
rdm/Gdm × ey .
B
However, the center-of-mass corollary requires that the integral in parentheses be zero
(see Eq. (9.3)). Thus for any rigid body falling in a uniform gravitational field, the
external moment about the center of mass must be zero:
MG = 0.
Note that this is not a general principle of dynamics but rather a consequence of the
specific force field acting on the body. In fact, for the more general 1/r 2 gravitational
field, there is a moment about the center of mass of the falling body. We explore this
fact in Tutorial 9.2.
Example 9.5 The Moment on a Vibration-Isolation Table
Figure 9.5 shows a side view of a planar model of a vibration-isolation system
consisting of a heavy table of length L supported by two springy legs with spring
constants kA and kB . (We assume the springs stay vertical when the table rotates.)
This is a two-degree-of-freedom system described by the coordinates y (the height
of the center of mass) and θ (the rotation angle of the table). Both coordinates are
shown in Figure 9.5a. We use two frames, as shown in Figure 9.5b: inertial frame
I = (O, ex , ey , ez ) and body frame B = (G, b1, b2, b3). The transformation array
between the inertial frame and the body frame has the usual form:
ex
ey
b1
b2
cos θ
sin θ
− sin θ
cos θ .
Example 9.10 solves for the equations of motion of the table in translation and
rotation using the separation principle. Here we find the moment on the table about
its center of mass due to the discrete forces acting on it, as shown in Figure 9.5c.
The previous example showed that a uniform gravitational field can be modeled
as a single force acting at the center of mass while producing no moment about the
center of mass (independent of the shape of the rigid body). Thus the only moments
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B
θ
G
A
kB
y
kA
(a) Coordinates
B
b2
I
ey
b1
kA
kB
O
ex
(b) Reference frames
– sin θ冥 ey
k A冤y – yA – L
2
L sin θ e
–kB冤y – yB + –
冥y
2
–mGgey
(c) Free-body diagram
Figure 9.5 A vibration-isolation system consists of a table of length L supported by two stiff
springs.
about the center of mass of the table are produced by the spring forces FA and FB
acting at each end. To find the spring forces we need to know the displacement of
each spring as a function of the two coordinates, y and θ. Consider the position of
point A relative to O:
rA/O = rG/O + rA/G = yey −
= yey −
L
b1
2
L
L
cos θex − sin θey ,
2
2
where we have used the transformation table between the unit vectors of B and I .
The spring force at A is given by the displacement of the spring in the ey direction,
FA = −kA(y −
L
sin θ − yA)ey ,
2
where yA is the unstretched length of the spring at A. A similar analysis gives the
spring force at B,
DYNAMICS OF A PLANAR RIGID BODY
349
A
FB
Q
A
(ext)
FA
B
G FB
FA
G
B
(a)
(b)
G
(c)
Figure 9.6 A couple and pure torque acting on a rigid body about the center of mass G and
another point Q. (a) Couple about G. (b) Pure torque. (c) Couple about Q.
FB = −kB (y +
L
sin θ − yB )ey ,
2
where yB is the unstretched length of the spring at B.
The moment about G is found using Eq. (9.11):
MG = rA/G × FA + rB/G × FB
L
L
L
L
= − b1 × −kA(y − sin θ − yA)ey + b1 × −kB (y + sin θ − yB )ey
2
2
2
2
L
L
L
L
= kA cos θ (y − sin θ − yA)ez − kB cos θ(y + sin θ − yB )ez . (9.13)
2
2
2
2
Example 9.6 The Moment of a Couple
As described in the text, the moment about the center of mass in Euler’s second law
arises either from a collection of contact forces acting on the body at discrete points
or from a nonuniform field force. Among the set of contact forces we consider, there
is a special set that warrants highlighting. These are forces that produce a moment on
the rigid body, and thus rotational motion, but no net force and thus no translational
acceleration. We call such a set of forces a couple. Figure 9.6a illustrates a couple
acting on a rigid body. A couple consists of two forces, FA and FB = −FA, of equal
magnitude F and opposite directions acting on the rigid body at two collinear points
equidistant from the center of mass. The net force on the rigid body is thus zero:
FG = FA + FB = 0.
Without loss of generality, assume the forces act perpendicular to the line connecting
A and B.
The moment of the couple, however, is not zero. Let b be the unit vector directed
from B to A, and let F̂A be the unit vector in the direction of FA. Using Eq. (9.11),
the moment is
MG = rA/G × FA + rB/G × FB
= rb × F F̂A + (−rb) × (−F F̂A) = 2rF b × F̂A = 2rF b3,
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CHAPTER NINE
Figure 9.7 Pure torque.
where b3 points into the page and r is the distance from G to each of the two points
of application, A and B. Thus a couple produces zero total force and a moment of
magnitude 2rF about the center of mass. Note that a couple can consist of any number
of forces equally distant from the center of mass, as long as their vector sum is zero. For
n forces around the center of mass, the resulting moment has magnitude MG = nrF .
It is common to consider the limiting case where the distances of the applied forces
in a couple go to zero while the magnitude of the forces increases such that the product
remains constant. The resulting moment about G is called a pure torque and written
as τ. Figure 9.6b illustrates a pure torque. This is, of course, an abstraction, as all
moments, if you examine them with enough fidelity, consist of a number of applied
forces. However, the pure-torque concept can be very convenient for many problems.
Examples of pure torques abound, including screwdrivers, electric drills, and motors
of all types. Even twisting off the lid of a jar is an example of a pure torque (see
Figure 9.7).
The most important property of couples and pure torques is that they are independent of the reference point. The moment on the rigid body is the same no matter where
the couple is applied.1 This property is actually fairly easy to show. Consider the same
couple but now equidistant from some other point Q = G, as shown in Figure 9.6c.
The moment about Q is (see Section 7.3)
MQ = rA/Q × FA + rB/Q × FB
= rb × F F̂A + (−rb) × (−F F̂A) = 2rF b × F̂A.
What is the moment due to the couple about the center of mass? It is
MG = rA/G × FA + rB/G × FB ,
where here the two vectors rA/G and rB/G are no longer equal and opposite. However,
using the usual vector triads, we can rewrite the moment:
1 This is in contrast to the moment due to a single force: the moment about an arbitrary point Q is generally
different than the moment about the center of mass G.
DYNAMICS OF A PLANAR RIGID BODY
351
MG = (rQ/G + rA/Q) × FA + (rQ/G + rB/Q) × FB
= (rQ/G + rb) × (F F̂A) + (rQ/G − rb) × (−F F̂A) = 2rF b × F̂A = MQ.
The moment of the couple about the center of mass is exactly the same as the moment
about Q. Thus pure torques or couples can be applied anywhere on a rigid body and
produce the same moment with no translational force.
9.3.2 The Moment of Inertia
We use Eq. (9.10) in almost all our analysis of rigid-body motion, both here and in
Chapter 11, when we study three-dimensional motion. To make it useful, however,
we need an expression that relates the angular momentum about the center of mass
to the scalar coordinate for orientation, θ, so we can convert Euler’s second law to a
scalar equation of motion. To do so requires some manipulation.
We begin with the expression for the angular momentum in Eq. (9.8),
I
hG =
N
mi ri/G × Ivi/G.
i=1
We again use the discrete form in this discussion for simplicity; the integral form
for a continuous rigid body follows a similar set of steps. We next use the transport
equation (Definition 8.1) to rewrite the inertial velocity of each particle in terms of
the angular velocity of the rigid body:
I
vi/G = Bvi/G +IωB × ri/G.
(9.14)
=0
The body-frame velocity is zero because the body is rigid. At this stage, the specialization to a rigid body comes into play. If the collection of particles is rigid, then
by definition, the particles have no velocity in the body frame relative to the center of mass (see Section 8.1). The result is that the angular momentum about the
center of mass is
I
hG =
N
mi ri/G × (IωB × ri/G).
(9.15)
i=1
We simplify the following calculations by using the constraint that the rigid body
only rotates in the plane. That is, IωB = IωB b3, which yields
I
hG =
N
mi IωB ri/G × (b3 × ri/G).
(9.16)
i=1
We also assume that the body itself is confined to the plane, so that ri/G = xi b1 + yi b2
(in Cartesian coordinates). Thus the vector triple cross product in Eq. (9.16) results
in a vector in the b3 direction (try it!), implying that the angular momentum is also in
the ±b3 direction. The angular momentum is thus IhG = hGb3.
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CHAPTER NINE
Given the direction of the angular momentum, we need to find an expression for
its magnitude, hG = IhG. We do so by reversing the order of both cross products
and then taking the dot product of Eq. (9.16) with b3:
hG = IhG . b3 =
N
mi IωB (ri/G × b3) × ri/G . b3.
i=1
Note that we have put parentheses around the first cross product ri/G × b3 because,
in the next step, we treat this cross product as a single vector. We now use the scalar
triple product identity (a × b) . c = a . (b × c) (see Appendix B) to find
hG = IωB
N
mi ri/G × b3 . ri/G × b3 .
i=1
Since b3 is always perpendicular to ri/G in this planar situation, we can use the
definition of the cross product to show that
ri/G × b3 . ri/G × b3 = (xi b1 + yi b2) × b32 = xi2 + yi2 = ri/G2.
This expression makes the magnitude of the angular momentum of the rigid body
about G equal to
I B
hG = ω
N
mi ri/G2 .
i=1
(9.17)
=IG
Eq. (9.17) leads us to the next definition.
Definition 9.2 The moment of inertia IG of a planar rigid body about its center of
mass G is the sum of the mass-weighted squared distance of each point on the
body to the center of mass:
IG =
N
mi ri/G2.
i=1
For a continuous rigid body B with density ρ(rdm/G), we have
IG =
B
rdm/G2dm =
B
rdm/G2ρ(rdm/G)dV .
We practice calculating the planar moment of inertia in the examples below. Note
that the planar moment of inertia is a scalar quantity.
With the introduction of the moment of inertia, Eq. (9.17) simplifies to hG =
IGIωB . The angular momentum of the rigid body B with respect to absolute space
can thus be compactly written as
DYNAMICS OF A PLANAR RIGID BODY
353
hG = IGIωB = IGIωB b3.
I
(9.18)
In the next section we use this equation to find the rotational equations of motion of
a planar rigid body.
Example 9.7 Calculating the Planar Moment of Inertia
(Cartesian Coordinates)
This example uses Cartesian coordinates to compute the planar moment of inertia
about the center of mass of the square rigid body shown in Figure 9.8a. To use
Definition 9.2, we first specify the position rdm/G = xb1 + yb2 and volume (actually
area) element dV = dxdy. Note that we express the position as components in the
body frame B = (G, b1, b2, b3). Assuming mass m and uniform density ρ, we have
IG = ρ
l
l
−l
−l
8
(x 2 + y 2)dxdy = ρl 4.
3
Using ρ = m/(4l 2), we have IG = 2ml 2/3.
Example 9.8 Calculating the Planar Moment of Inertia
(Polar Coordinates)
We now use polar coordinates to compute the moment of inertia of the circular rigid
body shown in Figure 9.8b. The position of a small mass element dm with respect to
the center of mass G is rdm/G = r cos θ b1 + r sin θb2 and the volume (area) element
is dV = rdrdθ. Assuming mass m and uniform density ρ, Definition 9.2 yields
2π
IG = ρ
R
0
(r 2 cos2 θ + r 2 sin2 θ)rdrdθ =
0
π
ρR 4.
2
Using ρ = m/(π R 2), we have IG = mR 2/2.
B
B
R
l
G
G
l
l
l
(a)
(b)
Figure 9.8 Calculating the planar moment of inertia. (a) Square. (b) Circle.
354
CHAPTER NINE
Appendix D supplies the moments of inertia for a variety of common rigid body
shapes.
9.3.3 The Rotational Equation of Motion of a Planar Rigid Body
We can now use Eq. (9.18) for the angular momentum of a planar rigid body and
Euler’s second law to rewrite the rotational equation of motion in Eq. (9.10) in terms
of the moment of inertia and angular velocity:
Id
dt
I
hG =
Id
dt
IGIωB =
d
IG
dt
ω + IGIα B b3 = MG.
I B
The beauty of this formulation of the equations of motion for a rigid body—using
the separation principle and computing the moment of inertia as an integral of the
mass distribution in the body—is that the time derivative of the moment of inertia is
zero, which is seen from the definition:
d
d
mi ri/G2 = 0.
IG =
dt
dt
i=1
N
The moment of inertia is constant because, if the body is rigid, the distance of any
particle to the center of mass is constant. (Note that these derivatives have no frame
identified because we are differentiating scalar quantities.)
Thus we can write the equation of motion for the angular momentum about the
center of mass in its final form:
Id
dt
hG = IGIα B b3 = MG,
I
(9.19)
I B
ω̇ =
where we have also assumed that MG is in the b3 direction and have used Iα B =
θ̈. Note again that, as with a multiparticle system, when the total moment acting about
the center of mass is zero, then the law of conservation of angular momentum holds
and the angular momentum about the center of mass is constant. For the simple planar
situations considered in this chapter, that means the rigid body rotates at a constant
angular velocity IωB = θ̇.
Example 9.9 Disk Rolling down an Inclined Plane
This example determines the equation of motion for a rigid disk or wheel rolling
down an inclined plane, as shown in Figure 9.9. Recall from Example 8.2 that this
problem has only one degree of freedom, represented by coordinate θ in Figure 9.9a.
This is because of the no-slip condition, which tells us that the distance the disk travels
down the plane is entirely given by the amount the disk has rotated. This dependence
is usually represented by the differential relationship derived in Example 8.2 (see
Eq.(8.6)),
ẋ = −r θ̇,
(9.20)
where the coordinate x is the distance of the center of the disk from the bottom of the
incline, r is the radius of the disk, and θ increases clockwise.
DYNAMICS OF A PLANAR RIGID BODY
355
b2
θ
r
I
b1
G
P
B
e2
x
C
ey
ex
α
e1
O
(a) Coordinates
(b) Reference frames
Fex
Ney
–mge2
(c) Free-body diagram
Figure 9.9 Disk rolling down an incline of angle α.
We use the separation principle discussed in this section to derive the equation of
motion for θ. Figure 9.9b identifies three reference frames: an inertial frame I located
at O with its e1 direction along the ground, a second inertial frame C also located at
O but inclined so that ex is along the incline, and a body frame B fixed to the disk
and located at its center of mass. The transformation array between the unit vectors
of I and C is
e1
e2
ex
ey
cos α
− sin α
sin α
cos α .
The free-body diagram for the disk is shown in Figure 9.9c.
We begin by finding the translational equation of motion for the center of mass G
of the disk. From Euler’s first law, using the free-body diagram in Figure 9.9c and
Cartesian coordinates in C, this equation becomes
mẍex = F ex + N ey − mge2,
where we have assumed that the disk does not move in the ey direction and F is the
(static) friction force between the disk and the incline. Using the transformation array
yields the scalar equation of motion in the ex direction:
mẍ = −mg sin α + F.
By differentiating the no-slip condition in Eq. (9.20) we can substitute for ẍ to obtain
−mr θ̈ = −mg sin α + F.
(9.21)
To find F we use the rotational equation of motion for the disk in Eq. (9.19) with
angular velocity IωB = θ̇b3 and moment about G given by MG = rF b3:
IGθ̈ = rF,
(9.22)
356
CHAPTER NINE
where the moment of inertia of the disk about its center is given by IG = 21 mr 2 (see
Example 9.8). Eqs. (9.21) and (9.22) can be combined to eliminate F and obtain the
final equation of motion for the rolling disk,
3
mr θ̈ = mg sin α
2
or, when rearranged,
θ̈ =
2g sin α
.
3r
(9.23)
Example 9.10 Vibration Isolation
Example 9.5 introduced the vibration-isolation system consisting of a heavy table
supported by two springy legs. The coordinates, reference frames, and free-body
diagram are shown in Figure 9.5. In this example we find the equations of motion
for the table’s position and orientation.
Recall that the position of the table’s center of mass is described by the height y
and its orientation in the inertial frame by the rotation angle θ. The position of the
center of mass in I is thus
rG/O = yey ,
and the angular velocity of the table in the inertial frame is IωB = θ̇ez .
To find the equations of motion we separately use Euler’s first law on the center of
mass and then Euler’s second law in Eq. (9.19) for motion about the center of mass.
There are three forces acting on the table: gravity acting at the center of mass,
FG = −mGgey ,
and the two spring forces, FA and FB , acting at endpoints A and B, respectively.
Example 9.5 showed that
L
FA = −kA y − sin θ − yA ey
2
L
FB = −kB y + sin θ − yB ey ,
2
where yA and yB are the unstretched lengths of the springs.
The translational equation of motion for the center of mass is found from Euler’s
first law in the ey direction,
mGÿ + mGg + (kA + kB )y + (kB − kA)
L
sin θ = kAyA + kB yB .
2
(9.24)
The equation of motion for θ is found by substituting the moment expression in
Eq. (9.13) into Euler’s second law in Eq. (9.19):
DYNAMICS OF A PLANAR RIGID BODY
L
IGθ̈ + cos θ (kB − kA)y +
2
357
2
L
(kA + kB ) cos θ sin θ
2
(9.25)
L
= (kB yB − kAyA) cos θ.
2
Eqs. (9.24) and (9.25) are the equations of motion for the two degrees of freedom
of the table—translation and rotation.
Eqs. (9.24) and (9.25) also indicate the existence of a nonzero equilibrium. When
ÿ = θ̈ = 0, the equilibrium position of the table is given by the solution of two
equations:
y0 +
k y + k B yB − m G g
kB − kA L
sin θ0 = A A
kB + k A 2
kA + k B
y0 +
kA + kB L
k y − k A yA
sin θ0 = B B
.
kB − k A 2
kB − k A
The solution gives the equilibrium height and angle:
yB + yA mGy(kB + kA)
−
4LkAkB
2
yB − yA mGg(kB − kA)
θ0 = arcsin
+
.
L
2Lka kB
y0 =
As you might have expected, in equilibrium the springs are slightly compressed and
the table is lower than the unstretched lengths of the springs due to gravity, while it has
a slight tilt because of the unequal spring constants. If kA = kB , then the equilibrium
reduces to a small vertical sag in y and no rotation (θ0 = 0). The equations of motion
in y and θ also completely uncoupled. That is why it is important to find table legs
with equal spring constants!
Finally, we can make the small-angle approximation and rewrite the equations of
motion for the table for small motions about its equilibrium. Letting y = δy + y0 and
θ = δθ + θ0 in Eqs. (9.24) and (9.25) yields2
δ ÿ +
(kA + kB )
(k − kA) L
δθ = 0
δy + B
2
mG
mG
δ θ̈ +
k B − kA L
k + k B L2
δy + A
δθ = 0,
IG 2
IG
4
where δy and δθ are small deviations from the equilibrium values y0, θ0.
We discuss the solution of such coupled linear differential equations for vibrational
problems in Section 12.1. Note again that if the spring constants are equal, kA = kB ,
these separate into two simple harmonic motions. Section 12.1 also discusses how
such a system is used to isolate the table from the motion of the ground.
2 We
discuss linearization in Section 12.2.2.
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CHAPTER NINE
9.3.4 Impulsive Moments on a Rigid Body
Chapter 4 (Definition 4.4) defined the angular impulse on a particle P associated with
an applied moment to be
t2
MP/O (t1, t2) =
MP/O dt.
t1
We also solved for the integrated, impulsive form of the angular-momentum equation
of motion for a particle (Eq. (4.17)). We can do the same for the equation of motion
of the angular momentum of a rigid body in Eq. (9.10).
Suppose that an impulsive moment or torque is applied to a rigid body. Then
integrating Eq. (9.10) gives
I
hG(t2) = IhG(t1) + MG(t1, t2).
(9.26)
In the planar case, Eq. (9.26) can be simplified nicely using Eq. (9.18) to substitute
for the angular momentum:
I B
ω (t2) = IωB (t1) +
1
MG(t1, t2),
IG
(9.27)
where IωB = IωB b3.
Thus the application of an impulsive moment to a planar rigid body instantaneously
changes its angular velocity, just as an impulsive force instantaneously changes the
velocity of a particle. In fact, the force associated with the impulsive moment will
also cause the center of mass of the rigid body to instantaneously change its velocity,
as described in Chapter 4.
Example 9.11 Cue Shot
Part of the art of playing pool is knowing just where to hit the cue ball with the cue.
Hit it high and the cue ball rolls forward faster than the no-slip condition, making
it follow another ball after contact. Strike it low, as in Figure 9.10, and the cue ball
rolls too slowly and slides forward, making it bounce back after hitting another ball.
This example computes the height above the centerline of the cue ball at which an
impulsive strike by the cue makes it roll without slipping.
Figure 9.11 shows the model for the cue striking the cue ball a distance h above the
centerline. The cue ball is a sphere with radius r, mass m, and moment of inertia about
a central axis of IG = (2/5)mr 2 (see Appendix D). It is struck by a force Fc = Fc e1
over a short interval, resulting in an impulsive force Fc (t1, t2) in the e1 direction. If
the ball starts at rest at t = t1, then from Eq. (4.2), its change in velocity will be
mIvG/O (t2) = mv(t2)e1 = Fc (t1, t2) = Fc te1,
DYNAMICS OF A PLANAR RIGID BODY
359
Figure 9.10 Pool cue hitting the cue ball. Image courtesy of Shutterstock.
x
h
r
B
I
G
I
ω
b2
e2
Fce1
b1
B
e1
–mge2
–Ff e1
Ne2
(a) Coordinates
(b) Reference frames
(c) Free-body diagram
Figure 9.11 A pool ball being struck a distance h above its centerline.
where t2 is the time just after the impulse and t = t2 − t1. The impulse imparted by
the cue also creates an impulsive moment on the cue ball about its center of mass:
MG(t1, t2) = he2 × Fc = −hFc te3.
From Eq. (9.27) the angular velocity of the cue ball about its center of mass after
impact satisfies
IGIωB (t2)e3 = −hFc te3.
Substituting for the linear impulse from the translational equation gives
2
IGIωB (t2) = mr 2IωB (t2) = −hmv(t2).
5
(9.28)
For the ball to roll without slipping, it must satisfy the no-slip condition at t2,
v(t2) = −r IωB (t2).
Substituting this condition into Eq. (9.28) gives the required height above the center
for the cue ball to start to roll without sliding:
h=
2
r.
5
360
CHAPTER NINE
If h < (2/5)r, then the ball slides faster than it rolls and friction will retard the
horizontal motion, slowing it down. If h > (2/5)r, the ball rolls too fast for its
translational motion and sliding friction acts forward, increasing the linear velocity
while decreasing the angular velocity, so that the ball speeds up.
9.4 Rotation about an Arbitrary Body Point
There are many situations where we are interested in writing the angular momentum
about an arbitrary point of the body rather than about the center of mass, as in
Figure 9.12, even though angular momentum separation no longer holds. Section 7.3
discussed in detail the dynamics of a collection of particles relative to an arbitrary
point Q. The resulting equation of motion, given in Eq. (7.17), was
Id
hQ = MQ + rQ/G × mGIaQ/O ,
I
dt
where the angular momentum about Q is
I
hQ =
N
ri/Q × mi Ivi/Q
i=1
and we have dropped the (ext) from the moment.
Fortunately, the derivation in the previous section holds exactly for motion relative
to Q. If we also assume that Q is fixed to the body, so that Bvi/Q = 0 for every mass
particle, then use of the transport equation leads to
I
hQ = IωB b3
N
mi ri/Q2 = IQIωB b3.
i=1
(9.29)
=IQ
b2
B
I
vQ/O
b1
Q
i
I
Ꮾ
rQ/G
e2
ri/G
rQ/O
G
ri/O
I
ωB
rG/O
O
e1
e3
Figure 9.12 Planar rigid body B with body frame B attached to an arbitrary point Q.
DYNAMICS OF A PLANAR RIGID BODY
361
The moment of inertia about Q for a discrete collection of particles is
IQ =
N
mi ri/Q2,
i=1
and for a continuous rigid body B it is
IQ =
B
rdm/Q2ρ(rdm/O )drdm/O .
The equation of motion for the angular momentum about Q then reduces to
Id
dt
hQ = IQIα B b3 = MQ + rQ/G × mGIaQ/O .
I
(9.30)
Eq. (9.30) can be used to solve a rotational problem about an arbitrary point on
the body. It is most useful for problems involving multiple rigid bodies where the
attachment point is moving, thus making it easier to use Eq. (9.30) than to compute
the angular momentum about the center of mass. Many problems have one point of
the rigid body pinned in absolute space, so that Q is now an inertially fixed point, and
Eq. (9.30) reduces to Euler’s second law. The effort is in finding the moment of inertia
and moments relative to Q. Usually we already have a description of the inertia of
the rigid body and the moments acting on it relative to the center of mass, so it seems
like extra work to start from the beginning to compute the moment of inertia about Q
and total moment relative to Q. In fact, there are shortcuts that simplify finding these
quantities. In the next two subsections we introduce two well-known theorems for
making these calculations easier. A short discussion of the special case of a pinned
rigid body then follows.
9.4.1 The Moment Transport Theorem
Eq. (9.30) requires an expression for the external moment MQ on the rigid body about
Q. This quantity is obtained from the definition of the moment:
MQ =
N
ri/Q × Fi(ext).
(9.31)
i=1
Suppose, however, that we already have an expression for the moment about the center
of mass G from Eq. (9.11). It turns out that there is a simple formula for relating the
moment about Q to the moment about G without having to recompute the sum in
Eq. (9.31). We begin by replacing ri/Q in Eq. (9.31) with the difference of vectors
relative to O in Figure 9.12:
MQ =
N
i=1
(ri/O − rQ/O ) × Fi(ext).
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CHAPTER NINE
We next substitute for ri/O from the vector triad ri/O = ri/G + rG/O :
MQ =
N
(ri/G + rG/O − rQ/O ) × Fi(ext).
i=1
Distributing the terms in parentheses yields
MQ =
N
i=1
ri/G × Fi(ext) +
N
(rG/O − rQ/O ) × Fi(ext).
i=1
=MG
The last term can be simplified using the vector triad rQ/G + rG/O = rQ/O to obtain
MQ = MG − rQ/G ×
N
Fi(ext).
(9.32)
i=1
Eq. (9.32) is known as the moment transport theorem. It can be used to write the
moment about point Q in terms of the moment about G and the vector between Q and
G. We have chosen to use the center of mass here because it is a common reference
point due to the separation of angular momenta, but it applies to any pair of points
fixed to the body.
Example 9.12 The Gravitational Moment on a Compound Pendulum
Look back to Example 8.1 (and Figure 8.2), which considered the kinematics of
a compound pendulum. Soon we will derive the full equations of motion for the
compound pendulum. To do so, we need the moment acting on the rigid body relative
to the attachment point O. To find the moment relative to O, we could compute the
moment due to the gravity force on each mass element and integrate over the body.
However, the moment transport theorem makes finding the moment much easier. We
already showed (in Example 9.4) that the total moment about the center of mass of a
rigid body in a uniform gravitational field is zero:
MG = 0.
Thus for the compound pendulum, the moment transport theorem (Eq. (9.32)) in
continuous form gives (ignoring the force at O because it cancels)
MO = −rO/G ×
B
dmge1
= −mGgrO/G × e1.
This moment is the same as if we treated the compound pendulum as a single
particle at G and calculated the moment relative to O. As we discussed in Example 9.3,
treating the moment on the compound pendulum as if it were equivalent to the moment
on a simple pendulum located at the center of mass is not correct in a 1/r 2 gravitational
DYNAMICS OF A PLANAR RIGID BODY
363
field because the center of mass and center of gravity are not the same for most rigid
bodies.
9.4.2 The Parallel Axis Theorem
Recall that Eq. (7.16) in Section 7.3 expressed the angular momentum about point Q
in terms of the angular momentum about the center of mass:
hQ = IhG + mGrQ/G × IvQ/G.
I
(9.33)
This equation leads to an expression for the moment of inertia about Q in terms
of the moment of inertia about the center of mass, which in turn eliminates the need
to recalculate the moment of inertia for each new point. In Eq. (9.18) IhG is given in
terms of the moment of inertia about G. Look at the second term in Eq. (9.33) and
use the transport equation:
mGrQ/G × IvQ/G = mGrQ/G ×
B
vQ/G +IωB × rQ/G ,
=0
where the velocity of Q in the body frame is zero, since we are assuming that Q is
fixed to the rigid body. The result is the following triple cross product:
mGrQ/G × IvQ/G = mGrQ/G ×
I B
ω × rQ/G .
Just as in Section 9.3.2, we recognize that this triple cross product is always in the b3
direction for the planar case, leading to
mGrQ/G ×
ω × rQ/G = mGIωB rQ/G2b3.
I B
We now rewrite the expression for IhQ in Eq. (9.33) as
hQ = IGIωB b3 + mGIωB rQ/G2b3.
I
A comparison to Eq. (9.29) shows that the moment of inertia about Q can be written
IQ = IG + mGrQ/G2.
(9.34)
Eq. (9.34) is known as the parallel axis theorem. If the moment of inertia about an
axis through the center of mass is known, then this formula can be used to find the
moment of inertia about any other parallel axis in terms of the perpendicular distance
between them.
Example 9.13 The Moment of Inertia of a Compound Pendulum
Examples 8.1 and 9.12 looked at a compound pendulum of arbitrary shape. In the
next section we solve for the equation of motion of the compound pendulum and find
that it depends on the moment of inertia about the pinned point O. Here we use the
parallel axis theorem to find the moment of inertia for a specific pendulum geometry.
364
CHAPTER NINE
O
l
Gr
a
Gd
R
(a)
(b)
Figure 9.13 (a) Pendulum clock. (b) Compound pendulum model. Image (a) courtesy of
Shutterstock.
A pendulum clock consists of a long rod of length l and width a with a circular
disk of radius R at its end, as shown in Figure 9.13b. The center of mass Gr of the
rod is located at its midpoint, and the center of mass Gd of the disk is at its center.
From Example 9.8, the moment of inertia of the disk is
1
IGd = md R 2.
2
From Appendix D, the moment of inertia of the rod about its center of mass is
IGr =
mr 2
(l + a 2).
12
Using the parallel axis theorem, the moment of inertia about the attachment point
O is the sum IO = IOd + IOr :
1
m
l2
IO = md R 2 + md (l + R)2 + r (l 2 + a 2) + mr
2
12
4
2
2 R
3 R
1
1 a
2
2
= md l 1 + 2
+
+ mr l 1 +
.
l
2 l
3
4 l
For a rod that is long and thin compared to the disk radius this is often approximated
by
1
IO ≈ (md + mr )l 2.
3
DYNAMICS OF A PLANAR RIGID BODY
365
This example also shows how the moment of inertia of a more complex object can
be found by piecing together the moments of inertia of its parts and using the parallel
axis theorem.
Example 9.14 Disk Rolling down Incline, Revisited
Example 9.9 computed the equation of motion for a disk or wheel rolling down an
inclined plane by using the separation principle and finding equations for translation
of the center of the disk and rotation of the disk. The no-slip condition was used to
combine these into a single equation of motion for the angle θ of the disk. Here we
recompute the equation of motion by using Eq. (9.30) about the contact point P .
The angular momentum of the disk about P is given by Eq. (9.29),
I
hP = IP θ̇b3,
where we used the fact that IωB = θ̇. The moment of inertia of the disk about P is
found from the parallel axis theorem to be
3
1
IP = mr 2 + mr 2 = mr 2.
2
2
From Example 8.2, the no-slip condition states that the point of contact of the disk
and the ground is instantaneously fixed, implying that IaP/O = 0 in Eq. (9.30). Thus
the moment equation is:
Id
I
hP = IP θ̈b3 = mgr sin αb3,
dt
which results in the same equation of motion as in Eq. (9.23):
θ̈ =
2g sin α
.
3r
9.4.3 A Pinned Rigid Body
Example 9.14 was an example of the common situation where the point Q is fixed
in inertial space. In this case, the acceleration of Q relative to the inertial frame in
Eq. (9.30) is zero and the equation of motion for IhQ reduces to Euler’s second law,
Id
dt
hQ = IQIα B b3 = MQ.
I
(9.35)
Thus if we use the parallel axis theorem to find the moment of inertia about Q, the
equation of motion for the angular momentum about Q is the same as the equation
of motion about the center of mass G. We just have to be sure to take the moment of
the external forces about Q. This simplification is useful for many problems.
It is common practice, in particular for pinned problems, to introduce the definition
of a new quantity, the radius of gyration, which is used to describe the moment of
inertia of a body.
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CHAPTER NINE
b2
Ft b2
e2
O
e1
O
l
O
I b1
θ
G
(a) Coordinates
Fnb1
B
G
(b) Reference frames
mge1
G
(c) Free-body diagram
Figure 9.14 Compound pendulum consisting of a rigid body with moment of inertia IG pinned
at point O in the inertial frame.
Definition 9.3 The radius of gyration kQ relative to a point Q of a rigid body is
the equivalent distance from Q of a particle of mass mG needed to produce the
same moment of inertia as the rigid body about Q:
2 = IQ .
mG k Q
Thus the radius of gyration of a rigid body about its center of mass is kG =
IG/mG. By the parallel axis theorem, the following also holds for an axis through
a point Q a distance l away from the center of mass:
2
IQ = IG + mGl 2 = mG(kG
+ l 2).
Many handbooks list the radius of gyration rather than the moment of inertia; almost
always, this quantity is the radius of gyration relative to the center of mass, in which
case the subscript G is often dropped. The reason for using the radius of gyration
rather than the moment of inertia is that it is easier to measure experimentally, as seen
in the next example.
Example 9.15 The Compound Pendulum
This example returns to the compound pendulum of Examples 8.1 and 9.12. Our goal
now is to find the equation of motion for θ.
Since the body is pinned at point O (Figure 9.14) the easiest approach to this
problem is to balance the angular momentum and moment about O using Euler’s
second law for the total angular momentum in Eq. (9.35) with Q replaced by O. This
approach allows us to ignore the reaction forces at the pin in Figure 9.14c. We could
follow the same process as in Example 4.4 for the simple pendulum, but much of the
work has already been done. The angular momentum of the compound pendulum is
given by Eq. (9.29):
I
hO = IO IωB b3,
where IO is given by the parallel axis theorem: IO = IG + mGl 2. The moment about
O was found in Example 9.12 using the moment transport theorem:
MO = −mg(−lb1) × (cos θb1 − sin θb2) = −mgl sin θb3,
DYNAMICS OF A PLANAR RIGID BODY
367
where we still assume a uniform gravitational field. The resulting equation of motion
from Eq. (9.35) is
Id
dt
I
hO = IO θ̈b3 = −mgl sin θb3,
or more simply,
θ̈ +
mgl
IO
sin θ = 0.
(9.36)
Eq. (9.36) is the same equation of motion as the simple pendulum but with
frequency given in terms of the moment of inertia about the pinned point O. Thus
for small angles, the compound pendulum oscillates sinusoidally just as a simple
pendulum. Replacing the moment of inertia with the radius of gyration about O gives
θ̈ +
g
2 /l
kO
sin θ = 0.
(9.37)
Eq. (9.37) looks just like the equation of motion for a simple pendulum of length
2 / l. The radius of gyration is thus related to the length of the simple pendulum,
kO
with mass mG, of the same oscillation frequency. This is in fact how the moment of
inertia is measured for most rigid bodies. Rather than trying to calculate IG based on
the geometry of what is often a complex mass distribution, the body is simply hung
from a point and set oscillating. The oscillation frequency gives a direct measure of
the radius of gyration from Eq. (9.37) in terms of the distance to the center of mass, l.
Example 9.16 Pulling without Slipping
This example examines the slip-free motion of a torqued wheel, as shown in Figure 9.15a. The torque comes from a string that is tied around a spool attached to the
wheel’s axle. As shown in this example, the direction of motion of the wheel (left or
right) depends on the angle α of the rope.
Because the wheel is rolling without slipping, its translational dynamics are kinematically paired with its rotational dynamics; that is, we need only find one or the
I
ey
θ
T
T
G
–m Ggey
Ney
O
Q
(a)
T
α
rP/Q
Fex
P
P
rP/Q
ex
(b)
(c)
Figure 9.15 Pulling without slipping. (a) Free-body diagram. (b) rP/Q × T < 0. (c) rP/Q×
T > 0.
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other. We focus on deriving the rotational dynamics, for which we have at least two
options. The first is to relate the rate of change of the angular momentum about
the center of mass, IhG = IGIωB , to the total moment about the center of mass,
(ext)
, using
MG = N
i=1 ri/G × Fi
Id
dt
I
hG = M G .
The second option is to consider instead the time rate of change of the angular momentum about the wheel’s contact point Q with the ground. The angular momentum
(ext)
.
about Q is IhQ = IQIωB , and the total moment about Q is MQ = N
i=1 ri/Q × Fi
Examination of the free-body diagram in Figure 9.15a helps make a decision. The
first option has two torques and two unknowns, F and T (the tension in the string),
whereas the second option has only one unknown T due to the tension in the string
acting at point P . Pursuing the second option by using IhQ = IQθ̇b3 yields
Id
dt
I
hQ = IQθ̈b3 = MQ = rP/Q × T.
At this point, we know enough to determine the direction of the wheel’s motion
given the angle α. Consider the arrangements of the vectors rP/Q and T shown in
Figure 9.15b and c. For small α, the cross product rP/Q × T is negative, which implies
that the wheel rolls right (θ̈ < 0); for large α, the cross product is positive, which
implies that the wheel rolls left (θ̈ > 0). In the problems, you will have a chance to
determine the critical rope angle, α ∗, below which the wheel rolls right and above
which it rolls left.
9.5 Work and Energy of a Rigid Body
The penultimate task of this chapter is to specialize the treatment of work and energy
for multiparticle systems in Section 7.4 to a planar rigid body.
9.5.1 Kinetic Energy of a Rigid Body
As for a multiparticle system, we begin with the kinetic energy. Eq. (7.23) showed
that the kinetic energy of a collection of particles separates into the kinetic energy
due to motion of the center of mass and the kinetic energy due to motion about the
center of mass:
TO = TG/O + TG.
(9.38)
We can specialize the second term—the kinetic energy due to motion about the center
of mass—to rigid bodies. It is given in Eq. (7.25) for a general collection of particles:
TG =
N
N
1 I
1
mi vi/G . Ivi/G =
mi Ivi/G2.
2 i=1
2 i=1
DYNAMICS OF A PLANAR RIGID BODY
369
In a rigid body there is no movement of the individual particles relative to the
center of mass in the body frame. We thus use the transport equation on Ivi/G, as
when developing angular momentum in Eq. (9.14). Letting the body-fixed velocity
be zero allows us to rewrite the kinetic energy for motion about the center of mass:
TG =
N
N
1
1
mi IωB × ri/G2 = (IωB )2
mi b3 × ri/G2 .
2 i=1
2
i=1
=ri/G 2
This summation is just the definition of the moment of inertia about G! Thus we
can write a simple expression for the kinetic energy of a rigid body rotating about its
center of mass:
1
TG = IG
2
1
ω . IωB = IGIωB 2.
2
I B
(9.39)
The moment of inertia and angular velocity supply a very simple way for calculating
the total kinetic energy of the rigid body about the center of mass. We can also
substitute the expression for the angular momentum of a rigid body about its center of
mass from Eq. (9.18) into Eq. (9.39) to formulate another commonly used expression
for the kinetic energy of rotation:
TG =
1I . I B
hG ω .
2
(9.40)
Example 9.17 A Flywheel
Flywheels are a way of storing energy using only mechanical components (unlike
batteries, which use chemical storage). This is done by building up a high level of
rotational kinetic energy in a rotor with a large moment of inertia, such as the one
shown in Figure 9.16. The schematic shows a uniform disk of radius r rotating at rate
θ̇. Eq. (9.39) allows us to write the kinetic energy of the flywheel simply as
1
TG = IGθ̇ 2.
2
θ
r
G
Figure 9.16 Flywheel schematic.
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Since the moment of inertia about the center of mass of a uniform disk is given by
mr 2/2, this expression becomes
1
TG = mr 2θ̇ 2.
4
Applying this equation, we find that a 1 kg flywheel with a diameter of 1 m, rotating
at 100 rpm (about 10.5 rad/s), stores almost 7 J of energy. For comparison, a typical
flywheel used for electrical-power backup would have mass of 500 kg, be half a meter
in diameter, and rotate at 30,000 rpm—that’s more than 77 MJ!
We can also find the kinetic energy for rotation relative to an arbitrary point Q on
the body, as done in Section 9.4 for angular momentum. We start by writing the total
kinetic energy of the rigid body relative to the origin of the inertial frame:
TO = TG/O + TG
1
1
= mGIvG/O 2 + IGIωB 2.
2
2
We then replace IvG/O in the first term with the usual vector triad, IvQ/O = IvG/O +
Q/G , relating the velocity of Q to the velocity of G:
Iv
TO =
1
mGIvQ/O 2 −mG
2
1
vQ/O . IvQ/G + mGIvQ/G2
2
I
=TQ/O
1
+ IGIωB 2,
2
1
I
2
where TQ/O =
2 mG vQ/O is the kinetic energy of an equivalent particle of mass
mG located at point Q. Next we use the transport equation IvQ/G = BvQ/G + IωB ×
rQ/G = IωB × rQ/G to replace IvQ/G (Q is fixed to the rigid body):
TO = TQ/O − mGIvQ/O .
1
1
ω × rQ/G + mGIωB × rQ/G2 + IGIωB 2.
2
2
I B
= 21 mG (IωB )2 rQ/G 2
The last two terms should look familiar from our discussion of the parallel axis
theorem. Using the same planar rotation assumptions as in Section 9.4.2, we can
rewrite this equation in terms of the moment of inertia about Q and the kinetic energy
about Q:
TO = TQ/O + TQ − mGIvQ/O .
I B
ω × rQ/G ,
where
1
1
TQ = (mGrQ/G2 + IG)IωB 2 = IQIωB 2
2
2
(9.41)
DYNAMICS OF A PLANAR RIGID BODY
371
is the kinetic energy associated with rotation of the rigid body about Q. IQ is the
moment of inertia about point Q given by the parallel axis theorem in Eq. (9.34).
Note that Eq. (9.41) looks a lot like the separated expression for the kinetic
energy in Eq. (9.38) except for a cross term between the rotation of the rigid body
and the translation of point Q. As expected, complete separation only occurs when
considering rotation about the center of mass. However, for the special case of a pinned
rigid body (such as the compound pendulum), the inertial velocity of Q is zero, and
the rotational kinetic energy about Q is easily calculated using the angular velocity
of the rigid body and the parallel axis theorem.
9.5.2 Rigid Body Work
As in Chapter 7, we turn next to the work–kinetic-energy formula for a rigid body.
Since a rigid body is a collection of particles (a fact used throughout this chapter), the
work–kinetic-energy formula in Eq. (7.26) still holds:
W (tot) = TO (t2) − TO (t1),
where W (tot) is short for W (tot)({ri/O ; γi }N
i=1), the total external work on the body.
How might we further simplify this formula when specializing it to a rigid collection of particles? The first obvious simplification is to eliminate the internal work term.
Since the body is rigid, the constituent particles cannot move relative to one another,
and thus no internal work can be done over a trajectory. The work–kinetic-energy
formula thus immediately reduces to
W = TG/O + TG,
(9.42)
where we have used the separation property developed previously to separate the
TO (t2) − TO (t1) into the change of kinetic
change in total kinetic energy TO =
energy of the center of mass TG/O = TG/O (t2) − TG/O (t1) plus the change in kinetic
energy about the center of mass TG =
TG(t2) − TG(t1). We have also dropped the
(tot) superscript because there is no internal work.
Recall that Chapter 7 showed that the power associated with motion of the center
of mass was equal to the change in kinetic energy of the center of mass and that this
kinetic energy is therefore conserved if the total external force on the system is zero
(see Eq. (7.30)). This result remains very useful and important even for rigid bodies.
However, for a rigid body we can go a bit further. It is helpful to separate the total work
into the work associated with the motion of the center of mass and that associated with
the motion of the particles relative to the center of mass.
To begin, recall the definition of the external work on a collection of particles. In
the absence of internal work (i.e., for a rigid body), we have
W ({ri/O ; γi }N
i=1) =
N
i=1
γi
Fi(ext) . Idri/O .
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Next we rewrite the work in the velocity form of Eq. (5.1):
W ({ri/O ; γi }N
i=1) =
N
i=1
t2
t1
Fi(ext) . Ivi/O dt.
We now use a vector triad to replace the velocity with respect to the origin with motion
relative to the center of mass:
W ({ri/O ; γi }N
i=1) =
N
i=1
t2
t1
Fi(ext) . IvG/O dt +
N
i=1
t2
Fi(ext) . Ivi/Gdt. (9.43)
t1
=WG/O (rG/O ,θ;γG ,γθ )
WG/O (rG/O , θ; γG, γθ ) is the work done by the vector sum of all forces on the body
to move the center of mass along the trajectory γG while the rigid body rotates along
a rotational trajectory γθ . Exchanging the integration and summation operators yields
WG/O (rG/O , θ; γG, γθ ) =
t2
FG . IvG/O dt.
(9.44)
t1
Recall again how Chapter 7 showed that the rate of change of the kinetic energy
of the center of mass of a collection of particles was equal to the power acting on the
center of mass, or the derivative of the work in Eq. (9.44). Thus we can identify
the work in Eq. (9.44) with the change in kinetic energy associated with motion of
the center of mass:
WG/O = TG/O .
(9.45)
Again, if the total external force on the center of mass is zero, the translational kinetic
energy of the center of mass of a rigid body is conserved.
Also of interest is the simplification of the second (relative-motion) term for rigid
bodies in Eq. (9.43). Using the transport equation and recognizing that the particles
do not move relative to the center of mass in the body frame allows us to make the
following substitution:
W ({ri/O ; γi }N
i=1) = WG/O (rG/O , θ; γG , γθ ) +
N
i=1
= WG/O (rG/O , θ; γG, γθ ) +
t2
t1
N
t2 t1
i=1
Fi(ext) . (IωB × ri/G)dt
(ri/G × Fi(ext)) .IωB dt,
=MG
where we used the scalar triple product to rearrange terms (see Section B.4.2). The
summation over the cross product in the second term is the total external moment MG
acting on the rigid body! Thus we can write the total work done over the trajectory of
the rigid body in the simple form
DYNAMICS OF A PLANAR RIGID BODY
373
W = WG/O + WG,
where the work WG = WG(rG/O , θ; γG, γθ ) corresponding to motion of the rigid body
about the center of mass is
WG(rG/O , θ; γG, γθ ) =
t2
MG . IωB dt.
(9.46)
t1
The corresponding power associated with rotation about the center of mass is
PG = MG . IωB .
Note that just as the work on the center of mass can depend on orientation, the work
of motion about the center of mass possibly depends on the position of the center of
mass, since the moment MG can in general depend on both position and orientation.
(Tutorial 9.2 illustrates this dependence.)
Our previous result identifying the change in kinetic energy of the center of mass
to the work done on the center of mass lets us identify this second term as the change
in kinetic energy of rotation about the center of mass,
WG = TG.
(9.47)
Thus, as for the kinetic energy of the center of mass, if the total external moment
about G is zero (even if there are external forces), the rotational kinetic energy of a
rigid body is conserved. Note that the moment MG in Eq. (9.46) includes all external
moments applied to the rigid body, even couples and pure torques.
9.5.3 Potential Energy and Total Energy of a Planar Rigid Body
Our final step in this section is to replace the work term by a change in potential energy
when the applied forces (and torques) are conservative. Recall that Chapter 7 replaced
the internal and external work terms (when the forces on the various particles were
conservative) by a change in internal and external potentials for each particle. Since
for rigid bodies there is no change in internal potential energy, we can define the total
energy just as in Chapter 7:
EO (t) =
TG/O (t) + TG(t) + UO (t).
The total potential is due to the (external) conservative forces acting on the rigid body:
UO (t) =
N
i=1
Ui/O (ri/O (t)).
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In this equation, Ui/O (ri/O (t)) is the potential energy associated with each discrete
force acting at point i of the rigid body. For field forces, where every mass element of
the body contributes to the potential, we add an integral over the body B and replace
the potential energy with the potential per unit mass Udm/O (rdm/O ):
UO (t) =
B
Udm/O (rdm/O )dm.
(9.48)
As in Chapter 7, the total energy can be used in the work-energy formula leading
to Eq. (7.35), where the change in total energy is equal to the total nonconservative
work:
(nc)
+ WG(nc).
EO (t2) = EO (t1) + WG/O
Of course, conservation of energy holds for a rigid body just as for a multiparticle
system when no nonconservative work is being done.
Example 9.18 The Energy of a Compound Pendulum
Recall that Example 5.12 solved for the motion of a simple pendulum using energy
considerations. This example does the same for the compound pendulum of Example 9.15. Because the compound pendulum is pinned at O, it is easiest to find the
kinetic energy using the results for an arbitrary point (Eq. (9.41)) with Q = O:
1
TO = IO θ̇ 2,
2
where we have used the fact that the pinned point is not moving. As in Example 9.15,
the moment of inertia about O is equal to IG + mGl 2 by the parallel axis theorem.
From our result in Example 9.3 we know that the potential energy of the compound
pendulum in a uniform gravitational field must be mGg times the height of the center
of mass (see Tutorial 9.2 for a rigid body in a 1/r 2 gravitational field):
UO = −mGgl cos θ,
where the zero of potential energy is where θ = π/2.
The total energy of the compound pendulum relative to the attachment point O is
thus
1
EO = IO θ̇ 2 − mGgl cos θ,
2
(9.49)
where we used the fact that no nonconservative forces are acting on the pendulum.
(Remember that the constraint forces at the attachment point do no work.)
At this point, the problem is identical to Example 5.12 with mP l 2 replaced by
IO . For instance, we can use Eq. (9.49) to find the angular rate of the swing at any
angle given an initial height (or, equivalently, EO (0)). We can also use Eq. (9.49) to
integrate and solve for the trajectory in terms of the initial energy, as in Eq. (5.18).
DYNAMICS OF A PLANAR RIGID BODY
375
Example 9.19 The Energy of a Disk Rolling down an Inclined Plane
This example revisits the disk rolling down the inclined plane of Example 9.9 and
Figure 9.9a. Rather than find the equations of motion, however, we instead solve for
the total energy of the rolling disk. We determine how fast the disk is traveling at the
bottom of the plane, assuming it starts rolling some distance x(t0) up the plane.
Since we are assuming a uniform gravitational field, there is no potential energy
associated with the orientation of the disk, only the height of the center of mass:
UO = mGgx sin α.
The kinetic energy of the disk is given by the sum of the translational kinetic energy
of the center of mass and the rotational kinetic energy. Again using the inertial frame
C as in Example 9.9, the velocity of the center of mass is
C
vG/O = −ẋex .
The total kinetic energy is then
1
1
TO = TG/O + TG = mGẋ 2 + IGθ̇ 2.
2
2
Using the no-slip condition, ẋ = −r θ̇, and the disk inertia, IG = 21 mGr 2, we can write
the total energy of the disk as
3
EO = mGẋ 2 + mGgx sin α.
4
Is energy conserved in this problem? There are three forces acting on the disk:
gravity, the normal force due to the incline, and the friction between the disk and the
incline. We have included the conservative gravitational force in the potential above.
The normal force is a constraint force and does no work. What about friction? Because
the disk is rolling and not slipping, the friction force is always what it needs to be to
make the velocity of the contact point relative to the incline zero. In other words, it is a
constraint force; it reduces the degrees of freedom from two (rotation and translation)
to one, where the constraint equation is the no-slip condition. It thus does no work,
which implies the total energy is conserved.
We can thus find the final velocity of the disk, given an initial distance up the
incline, by simply equating the energies at the two points:
3
mGgx(t0) sin α = mGẋ(tf )2
4
or
ẋ(tf ) =
4
gx(t0) sin α.
3
Note the difference from the result for a sliding particle, which would be ẋ(tf ) =
2gx(t0) sin α. The key idea here is that we need to include the kinetic energy of
both translation and rotation of the disk in the total energy.
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9.6 A Collection of Rigid Bodies and Particles
We now turn to a general system in two dimensions consisting of interconnected
particles and rigid bodies. Remarkably, we have all the tools in place to solve for the
motion of an arbitrary system. Recall that Section 2.3 discussed the art of modeling a
physical system. Every dynamical system analyzed should be reduced to a collection
of particles, rigid bodies, and perhaps massless rods or strings (or other constraining
elements). The previous chapters of the book showed how to find the equations of
motion for a particle and for a multiparticle system. This chapter demonstrated how
to find the equations of motion for the translation and orientation (in the plane) of
a rigid body. Combining these methods enables us to solve for the motion of any
system.
The steps involved are actually rather straightforward. Once the system under
study is modeled as a collection of connected particles and rigid bodies (actually, this
is the hardest part), we simply break each connection and add constraint forces or
torques between the elements. We then find the equations of motion separately for
each of the constituent elements (particles or rigid bodies) using Newton’s or Euler’s
laws. Remember, particles moving in the plane have two degrees of freedom and thus
two equations of motion and rigid bodies have three degrees of freedom and thus
three equations of motion. Of course, we must include the constraint forces in
each of the equations of motion. (Don’t forget to draw the free-body diagrams!) Just
as we have done throughout the book, we then manipulate the vector equations of
motion (and constraint equations) to eliminate the constraint forces and find a scalar
equation of motion, in each coordinate, corresponding to each degree of freedom.3
To see that this method works, consider Np particles and Nr rigid bodies. They
may be connected through constraints or force-inducing elements, such as springs.
We know immediately from the results of Chapter 6 and this chapter that the center
of mass of the entire collection obeys Newton’s second law using the total external
force and that, if the total external force is zero, the center of mass is stationary or
moves at constant velocity. That is, the total linear momentum, which is the same as
the linear momentum of the center of mass of the entire collection, obeys Eq. (6.16):
N
I
pG/O =
p
I
mi vi/O +
Nr
mj IvGj/O
j =1
i=1
and
Id
dt
I
(ext)
pG/O = FG
,
where Ivi/O is the inertial velocity of particle i and IvGj/O is the inertial velocity of
the center of mass Gj of rigid body j .
3 It is worth noting here that the added complexity associated with introducing constraint forces and then
eliminating them by additional constraint equations is one of the motivations behind the Lagrangian method
described in Chapter 13. This advanced approach develops equations of motion for each coordinate (degree
of freedom) directly, without the need to introduce constraint forces.
DYNAMICS OF A PLANAR RIGID BODY
377
We can also look at the total angular momentum of the collection relative to some
point Q that may or may not be accelerating. Following the definition of total angular
momentum about an arbitrary point Q, we sum the individual angular momenta of
each particle and rigid body:
N
I
hQ =
p
I
hi/Q +
Nr
I
hj/Q,
j =1
i=1
where Ihi/Q is the angular momentum of particle i relative to Q and Ihj/Q = IhGj/Q +
Ih is the total angular momentum of rigid body j relative to Q. Note that we may
Gj
pick from a variety of choices for Q. It might be inertially fixed, it might be the center
of mass of the entire collection, it might be the center of mass of one of the rigid
bodies, or it might be one of the contact points.
We can now take the derivative of IhQ and distribute the derivative through each
sum. The derivative of the total angular momentum of the collection of particles is
equal to the external moment on it plus any (internal) moment from the rigid bodies.
Likewise, by Euler’s second law, the derivative of each of the rigid-body angular
momenta is equal to the external moment plus the (internal) moment from the particles
and other rigid bodies. However, if we again make the assumption that the internal
moments cancel (which will always be true if the only forces and moments acting
among the particles and rigid bodies are constraint forces), then we have Euler’s
second law for the entire collection:
Id
dt
I
hQ = MQ + rQ/Gp × mGp IaQ/O + mGr
Nr
rQ/j × IaQ/O ,
(9.50)
j =1
where Gp is the center of mass of the collection of particles, mGp is the total mass of the
particles, and mGr is the total mass of all the rigid bodies. While we can certainly use
Eq. (9.50) to try and find equations of motion, it is only slightly helpful for problems
with many degrees of freedom because it does not provide enough information to find
all the equations of motion. Its most important consequence is that, for an inertially
fixed point Q and zero external moment, the total angular momentum of a collection
of particles and rigid bodies is conserved. A classic example that uses conservation
of angular momentum of a system of rigid bodies is the satellite momentum wheel.
If a satellite encounters an impulsive disturbance that causes it to start to rotate (thus
gaining angular momentum), an internal wheel can be spun up such that the total
angular momentum stays fixed (only internal moments are generated), but the angular
momentum of the satellite goes to zero. Almost all low-earth orbiting satellites use
momentum wheels.
Finally, we can use the separation principle on each of the rigid bodies to rewrite
the total angular momentum as
N
I
hQ =
p
i=1
I
hi/Q +
Nr
j =1
I
hGj/Q +
Nr
I
hGj .
(9.51)
j =1
We can then examine the derivatives of each term in Eq. (9.51) just as in Chapter 7.
We would once again find Euler’s second law for the motion of each rigid body about
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CHAPTER NINE
its center of mass (and the motion of its center of mass relative to Q), only we would
have to be sure to include the internal constraint forces and moments acting among
the elements. That is exactly the process for finding equations of motion as described
at the beginning of the section.
We can perform a similar analysis for the energy of the collection of particles and
rigid bodies. Using the separation principle, the total energy is the sum of the kinetic
energy of all particles and rigid bodies plus the internal and external potential energies
of the system:
Nr
EO (t) = TGp/O (t) + TGp (t) +
Tj/O (t) + Tj (t) + UO(ext)(t) + UO(int)(t), (9.52)
j =1
where we have simplified a bit and considered only the energy relative to an inertially
fixed point O. The change in total energy across a trajectory is then given exactly by
Eq. (7.35); it is equal to the sum of the external and internal work done on the system.
Just as in Chapter 7, care must be taken to properly account for the internal forces
and internal work. Energy is not like angular momentum! Nevertheless, if the total
nonconservative work is zero, the total energy of the collection of rigid bodies and
particles is conserved.
Admittedly, this all sounds a bit complicated. It really isn’t. Studying the following
examples is a good way to see how to approach these more complex dynamics
problems.
Example 9.20 The Crane, Take 3
This example reexamines the overhead crane first discussed in Example 6.2. The
difference is that here instead of a simple pendulum hanging from the block (a point
mass at the end of a massless rod), there is a rigid arm, as in Figure 9.17a. Note that this
problem still has two degrees of freedom: the configuration is completely determined
by the horizontal position of the block and the angle of the arm. As usual, this implies
the presence of constraints and thus constraint forces. Since the arm is now a rigid
body, and thus nominally has three planar degrees of freedom (two translation and
one rotation), there must now be two constraint forces between the arm and the block
rather than the single tension force found for a simple pendulum. Figure 9.17c shows
these constraint forces, where the normal force N constraining the block to motion
only in x is still there. But now there is both a tangential reaction force and a normal
reaction force where the arm joins the block rather than just a tension force as with
a simple pendulum. (To solve this problem, we choose to express these forces as
components in the I frame.)
As described above, this problem can be solved by breaking apart the rigid bodies
and solving for the dynamics for each. The constraint forces can then be eliminated to
find the two equations of motion for x and θ. As in Example 6.2, I = (O, ex , ey , ez )
denotes the inertial frame, and B = (Q, er , eθ , ez ) denotes a body frame fixed to the
pendulum arm. The frames B and I are related by the transformation
er
eθ
ex
ey
sin θ
cos θ
− cos θ
sin θ .
(9.53)
DYNAMICS OF A PLANAR RIGID BODY
379
I
ey
eθ
g
x
B
Q
Q
O
ex
O
θ
P
2l
er
(a) Coordinates
(b) Reference frames
Ney
Fey
Q
–Tex
–mQgey
Q
Tex
–Fey
–mPgey
(c) Free-body diagrams
Figure 9.17 The crane, revisited.
The kinematics of block Q are the same as in Example 6.2:
rQ/O = xex
I
vQ/O = ẋex
I
aQ/O = ẍex .
Likewise, the kinematics of P , in this case the center of mass of the rigid rod, are
similar to those in Example 6.2:
rP/O = xex + ler
I
vP/O = ẋex + l θ̇eθ
I
aP/O = ẍex + l θ̈eθ − l θ̇ 2er ,
where l is half the length of the pendulum. Note that we have implicitly used a
constraint equation by changing to polar coordinates and setting r = l.
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We can now use Figure 9.17c and Euler’s first law to write the translational
equations of motion for the block Q and the center of mass P of the rigid rod:
(N + F − mQg)ey − T ex = mQẍex
(−F − mP g)ey + T ex = mP (ẍex + l θ̈eθ − l θ̇ 2er ).
(9.54)
(9.55)
Eqs. (9.54) and (9.55) represent four scalar equations, but there are now five unknown
quantities, ẍ, θ̈, N, F , and T . The fifth equation comes from considering the rotational
equation of motion for the rigid arm about its center of mass P :
Id
dt
I
hP = IP θ̈ez = (−ler ) × (−F ey ) − (ler × T ex )
= F lsin θez − T lcos θez .
These five equations can be manipulated to eliminate the constraint forces:
(mQ + mP )ẍ + mP l θ̈ cos θ − mP l θ̇ 2 sin θ = 0
(9.56)
(IP + mP l 2)θ̈ + mP gl sin θ + mP l ẍ cos θ = 0.
(9.57)
Finally, as done in Example 6.2, we can algebraically solve Eqs. (9.56) and (9.57)
for the equations of motion in x and θ:
3mP
3 mP
m
1−
cos2 θ ẍ −
g sin 2θ − P l θ̇ 2 sin θ = 0
mG
4mG
8 mG
3 mP
3 mP 2
3g
2
1−
cos θ θ̈ +
θ̇ sin 2θ +
sin θ = 0,
4 mG
8 mG
4l
where mG = mP + mQ and we used the fact that, for a thin rod, IP = 31 mP l 2. In the
limit of IP → 0, the equations of motion are identical to the equations found for the
point-mass crane.
Example 9.21 Particle on a Beam
A classic example of a combination rigid-body–particle system is a point mass sliding
on a pinned beam (or seesaw). This system is a common example that you might
explore in a course on automatic control. As shown in Figure 9.18a, the system
consists of a thin rigid beam, pinned to a support at its center of mass, and a particle,
which slides on the beam without friction. Note that this problem only has two degrees
of freedom and thus can be completely described using only two coordinates: θ, the
angle of the beam from horizontal, and x, the distance of the particle from the center
of the beam. We define an inertial frame I = (O, ex , ey , ez ) and body frame B =
(O, er , eθ , ez ), both attached to the center of the beam O, with the body frame fixed
to the beam (Figure 9.18b). The two frames are related by the transformation array
ex
er
eθ
cos θ
− sin θ
ey
sin θ
cos θ .
DYNAMICS OF A PLANAR RIGID BODY
381
I
B
eθ
x
P
P
g
θ
O
ey
er
O
ex
(a) Coordinates
(b) Reference frames
NPeθ
P
–mgey
Neθ
–NPeθ
O
–Mgey
(c) Free-body diagrams
Figure 9.18 Particle on a beam.
Figure 9.18c shows the free-body diagrams of the two components of the system.
Note that there are two normal forces involved here: N , the force of the beam support
on the beam, and NP , the force of the particle on the beam. Due to Newton’s third
law, NP must appear with opposite signs in both free-body diagrams. We start with
the particle P , whose kinematics are
rP/O = xer
I
vP/O = ẋer + x θ̇eθ
I
aP/O = (ẍ − x θ̇ 2)er + 2ẋ θ̇ + x θ̈ eθ .
Note that we have carried out the inertial derivatives using the body-frame coordinates.
Applying Newton’s second law and separating components in the er and eθ directions produces the following two equations for P :
m(ẍ − x θ̇ 2) = −mg sin θ
m(2ẋ θ̇ + x θ̈) = NP − mg cos θ.
(9.58)
(9.59)
We now have two equations of motion for the two degrees of freedom, but the problem
is not yet solved, since NP is unknown. To find it, we must consider the rotational
equation of motion of the beam. The beam is rotating about its center of mass O and
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the only force producing a moment about O is NP . All other forces in the free-body
diagram pass through O and contribute no moment. Thus the rotational equation of
motion is
Id
dt
I
hO = IO θ̈ez = xer × −NP eθ = −xNP ez ,
(9.60)
where IO is the moment of inertia of the beam about O. From Eq. (9.60), we find
NP = −
θ̈IO
,
x
which is valid as long as x = 0.
Substituting the expression for NP into Eq. (9.59) produces the equations of motion
for this system:
ẍ − x θ̇ 2 = −g sin θ
IO
2
θ̈ x +
= −gx cos θ − 2x ẋ θ̇.
m
Example 9.22 Yo-Yo De-Spin
In the early days of the space program a technique was developed to de-spin satellites,
that is, remove the satellite’s angular momentum, without the use of thrusters or fuel.
Dubbed “yo-yo de-spin,” it involved two masses connected to cords and wrapped
symmetrically around a cylindrical satellite. When released, the masses would unwind
and carry away the angular momentum due to the spin of the satellite. This device
is an excellent example of how a designer can use the conservation of total angular
momentum and total energy of a system of particles and rigid bodies. It is still in use
today.
Our planar model for the yo-yo device is shown in Figure 9.19. It consists of a
circular satellite of radius R and moment of inertia IG about its center of mass G.
It has an attached body frame B = (G, b1, b2, b3) and is rotating about its center of
mass with angular velocity
ω = IωB b3.
I B
There are two particles, P and Q, each of mass m, attached to cords that unwind from
the satellite. In Figure 9.19 they are shown after the cords have unwrapped a distance
l. Because of the symmetry, and equal masses, the center of mass of the entire system
is also at G, and the two particles unwind an equal length. We have also introduced a
polar frame C = (G, er , eθ , c3), where the radial unit vector er is directed toward the
contact point of the cord connected to P and the satellite. The angular velocity of C
in B is thus the usual:
B C
ω = θ̇c3.
We also assume that, at t = 0, the cords are completely wrapped around the satellite
and C is aligned with B. That is, θ(0) = 0. Thus the instantaneous cord length is
l = Rθ .
DYNAMICS OF A PLANAR RIGID BODY
383
B
b2
er
C
l = Rθ
eθ
P
θ
G
R
Q
b1
I
ωB
Figure 9.19 A yo-yo de-spin system consists of a cylindrical satellite and two masses, P and
Q, attached to cords wrapped around the satellite.
Because there are no external forces or moments on the system, G is the origin
of an inertial frame and the total angular momentum about G is conserved. The total
angular momentum consists of the sum of the angular momenta of the satellite and
the particles. From the kinematics of the problem, we have for particle P :
rP/G = Rer − Rθ eθ
vP/G = R IωC × er − R θ̇ eθ − Rθ IωC × eθ .
I
We use the addition property of angular velocities to find
ω = IωB + BωC
I C
= (IωB + θ̇)c3.
Substituting this expression into the previous one yields the inertial velocity of P :
I
vP/G = Rθ(IωB + θ̇)er + R IωB eθ .
Similarly, we have for Q:
rQ/G = −Rer + Rθ eθ
vQ/G = −Rθ(IωB + θ̇)er − R IωB eθ .
I
The (constant) total angular momentum of the system about G is
hG = IGIωB c3 + mrP/G × IvP/G + mrQ/G × IvQ/G,
I
which, after substituting for the position and velocities of P and Q, gives
I
hG = IGIωB + 2mR 2 IωB + θ 2(IωB + θ̇) c3.
Likewise, because there are no external or internal forces or moments, the total
energy about G is equal to the total kinetic energy and is also a constant of the motion.
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CHAPTER NINE
(The tension forces in the cords are constraint forces and therefore do no work.) The
total kinetic energy is found from Eq. (9.52) to be
1
1
1
EO = TG = IG(IωB )2 + mIvP/G2 + mIvQ/G2
2
2
2
1
= IG(IωB )2 + mR 2 (IωB )2 + θ 2(IωB + θ̇)2 .
2
Note that, because the total energy and angular momentum are conserved, they are
equal to their values at t = 0, which implies
I
hG = (IG + 2mR 2)Iω0B c3
1
EO = (IG + 2mR 2)(Iω0B )2,
2
where Iω0B = IωB (t = 0).
The expressions for the angular momentum and energy at t and t = 0 can be
equated to give the following two differential equations in the two unknowns θ and
IωB :
c(Iω0B − IωB ) = θ 2(IωB + θ̇)
c[(Iω0B )2 − (IωB )2] = θ 2(IωB + θ̇)2,
(9.61)
(9.62)
where
c=
IG
+ 1.
2mR 2
Dividing Eq. (9.62) by Eq. (9.61) gives
I B
ω0
+ IωB = IωB + θ̇,
which tells us that θ̇ = Iω0B = constant. The cords unwrap at a constant rate equal to
the initial spin rate of the satellite. Letting θ(t) = Iω0B t and substituting into Eq. (9.61)
gives the expression for the spin rate of the satellite as a function of time:
I B
ω
(t) = Iω0B
c − (Iω0B )2t 2
c + (Iω0B )2t 2
.
(9.63)
If we wish to completely de-spin the satellite, so that IωB (tf ) = 0, then we can use
Eq. (9.63) to find the final time tf for zero spin rate as a function of the initial spin
rate and mass properties of the satellite and particles:
√
tf =
c
IωB
0
.
DYNAMICS OF A PLANAR RIGID BODY
385
This expression can be used to find the length of cord lf necessary for complete
de-spin. Using l(t) = Rθ(t), we have
IG
I B
+ 1.
lf = Rθ(tf ) = R ω0 tf = R
2mR 2
9.7 Tutorials
Tutorial 9.1 The Center of Percussion
This tutorial revisits the compound pendulum of Example 9.15. We solve it here using
the separation of translational and angular momentum. Although this approach is
more complicated than that applied in Example 9.15 (because it involves introducing
and eliminating the reaction forces), it introduces the important concept of the center
of percussion.
An important observation is that, for a simple pendulum with only a point mass at
the end of a massless rod or string, the moment of inertia about the center of mass is
zero (it has no extent, after all), and the transverse reaction force is zero. This is the
case, though, only for the simple pendulum. For the more realistic physical pendulum,
there is always a transverse reaction force that must be accounted for in the dynamics.
Consider again the compound pendulum in Figure 9.14. Here we find the equation
of motion for θ by writing Euler’s first law for the center of mass and then balancing
angular momentum about the center of mass. The position of the center of mass,
expressed as components in the body frame B, is
rG/O = lb1.
The velocity is
I
vG/O =
Id
dt
(lb1) = l IωB × b1 = l θ̇b2
and the acceleration is
I
aG/O =
Id
dt
(l θ̇b2) = l θ̈b2 − l θ̇ 2b1.
Using the free-body diagram in Figure 9.14c, we can write Euler’s first law for the
center of mass,
Fnb1 + Ft b2 + mGg cos θb1 − mGg sin θb2 = mG(l θ̈b2 − l θ̇ 2b1),
(9.64)
where the necessary reaction forces at the pin have been included and we have used
the transformation array to express the gravitational force mGge1 in the pendulum
body frame. Eq. (9.64) is equivalent to the two scalar equations:
Fn + mGg cos θ = −mGl θ̇ 2
(9.65)
Ft − mGg sin θ = mGl θ̈.
(9.66)
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CHAPTER NINE
Since there are three unknowns—θ̈, Fn, and Ft —we need three equations. For the
third equation we use the time derivative of the angular momentum about the center
of mass, IhG = IGθ̇b3. The equation of motion for rotation about the center of mass is
given by Eq. (9.19). The moment about the center of mass comes from the transverse
reaction force at the pin,
MG = rO/G × Ft b2 = −Ft lb3.
The rotational equation of motion is
Id
I
dt
hG = IGθ̈b3 = −Ft lb3
or, in scalar form,
IGθ̈ = −Ft l.
(9.67)
Eqs. (9.65)–(9.67) provide three equations in the three unknowns Ft , Fn, and θ̈.
The equation of motion for θ̈ is found by solving Eq. (9.67) for Ft and substituting
into Eq. (9.66),
θ̈ +
mGgl
sin θ = 0.
IG + m G l 2
This equation is, of course, identical to the equation of motion found in Example 9.15
(Eq. (9.36)). However, this approach has given us an expression for each of the two
reaction forces:
Fn = −mG(g cos θ + l θ̇ 2)
Ft = −
mGgIG
IGθ̈
=
sin θ.
l
IG + m G l 2
We next explore a slight modification to the compound-pendulum problem. Suppose that an external force FAb2 is applied to the pendulum at point P a distance l
from the attachment point O (below the center of mass G), as shown in Figure 9.20.
This force is usually an impulsive force, but it need not be. In this case, the b2 translational equation in Eq. (9.66) is modified to become
Ft + FA − mGg sin θ = mGl θ̈.
(9.68)
The rotational equation is similarly modified. Here we find it easiest to return to the
angular momentum equation of motion about the pinned point O, as in Example 9.15,
but with an added moment due to the applied force (see Figure 9.20):
2
θ̈ = FAl − mGg sin θ.
IO θ̈ = mGkO
(9.69)
We make one further simplification and drop gravity from the problem. For instance, we may consider the pendulum to be rotating horizontally or that the force FA
is very large compared to gravity and short enough that gravity has little effect on the
DYNAMICS OF A PLANAR RIGID BODY
387
Ft b2
O
l′
Fnb1
G
mge1
P
FAb2
Figure 9.20 Free-body diagram of a compound pendulum with an applied force at point P a
distance l from attachment point O.
motion. In that case, Eqs. (9.68) and (9.69) can be combined to eliminate θ̈, resulting
in the ratio of transverse reaction force Ft to applied force FA:
2 /l
l − kO
Ft
l
− 1=
,
= 2
2 /l
FA k O / l
kO
(9.70)
2 / l is the
where kO is the radius of gyration defined in Definition 9.3. Recall that kO
length of the simple pendulum of mass mG with equal oscillation frequency. Eq. (9.70)
tells us something profound. If the external force is applied at a point P a distance
2 / l from the attachment point, there is no transverse reaction force! This point,
l = kO
the center of percussion, is easily found for any rigid body from the radius of gyration
and the mounting point. A good batter in baseball knows where to find the center
of percussion of a baseball bat; hitting a ball there feels effortless. Likewise, tennis
rackets are designed with a “sweet spot” at the center of the face so that a proper
swing will avoid a painful transverse reaction force.
Tutorial 9.2 A Dumbbell Satellite in a Circular Orbit
Chapters 4 and 7 presented examples that found the orbital motion of a satellite about
a large gravitating body, such as the earth. In them we treated the satellite as a point
mass acted on by a gravitational force given by Newton’s universal law of gravity. A
satellite is not, however, a point mass but an extended rigid body. It is thus acted on not
only by a gravitational force at its center of mass but also by a gravitational moment
about its center of mass due to the gradient of the gravitational field. In this tutorial
we calculate the gravitational moment on a simple model of a satellite consisting of
two point masses separated by a massless rod, as shown in Figure 9.21, and find its
equation of motion. We often call this system a dumbbell satellite. It is a reasonable
model of a satellite with a large central mass and an extended smaller mass, such as
an antenna or sensor. In fact, many satellites are intentionally designed this way, often
by including an extended mass on a long boom.
As in Example 4.9, we locate the center of mass G of the satellite using polar
coordinates (r, θ )I . We also introduce a polar frame C = (O, er , eθ , ez ), where er
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CHAPTER NINE
b1
m2
I
l2
r
ϕ
G
l1
m1
b2
e2
C
B
eθ
er
θ
O
O
(a) Coordinates
GmO m2
r2/O
–—
储r2/O储 3
I
e1
(b) Reference frames
GmO m1
r1/O
–—
储r1/O储 3
(c) Free-body diagram
Figure 9.21 Dumbbell satellite subject to a moment stemming from a gravitational gradient.
points from O to G. Thus the position of the center of mass in I can be expressed in
terms of components in C as
rG/O = rer .
In this example we are only interested in finding an expression for the external
moment about G acting on the satellite. In particular, we find an approximate formula
for the moment as a function of the satellite’s orientation in C. Figure 9.21a introduces
the rotational angle φ describing the rotation of the satellite relative to the er direction.
The transformation array between the body frame B = (G, b1, b2, b3) and polar frame
C is
er
eθ
b1
b2
cos φ
− sin φ
sin φ
cos φ .
We now write the moment on the satellite about G using the definition in Eq. (9.11),
the free-body diagram in Figure 9.21c, and the universal law of gravity:
GmO m2
GmO m1
MG = r1/G × −
(9.71)
r1/O + r2/G × −
r2/O .
r1/O 3
r2/O 3
We next substitute into Eq. (9.71) the two vector triads r1/O = rG/O + r1/G and
r2/O = rG/O + r2/G and use the fact that a vector crossed into itself is zero to get
MG = −
GmO m2
GmO m1
r2/G × rG/O . (9.72)
r1/G × rG/O −
rG/O + r2/G3
rG/O + r1/G3
From Figure 9.21b and the transformation array, we compute the two cross products,
r1/G × rG/O = −l1b1 × rer = rl1 sin φb3
r2/G × rG/O = l2b1 × rer = −rl2 sin φb3.
The expressions in the denominator are a bit more complicated. To simplify them
we use the fact that the satellite is much smaller than its distance from the center
of the earth: l1 r and l2 r. This fact allows us to approximate the moment by
DYNAMICS OF A PLANAR RIGID BODY
389
dropping any terms of order (l/r)2 and higher. We can expand the cubed magnitude
in the denominator in Eq. (9.72) as
rG/O + r1/G3 = (rG/O + r1/G) . (rG/O + r1/G)
= r 2 + 2rG/O . r1/G + l12
= r3 1 + 2
rG/O . r1/G
r2
+
3/2
3/2
l12
3/2
r2
.
We find a similar expression for the term involving r2/G. Since we are assuming
that l1/r and l2/r are small, we can use the binomial expansion in Appendix A
(Example A.1) to simplify the moment expression:
rG/O . r1/G
GmO m1
(l1 sin φ) 1 − 3
MG ≈ −
r2
r2
rG/O . r2/G
GmO m2
+
.
(l2 sin φ) 1 − 3
r2
r2
The two leading terms cancel by the center-of-mass corollary (Eq. (6.15)). Using the
transformation array, the two dot products are
rG/O . r1/G = −rl1 cos φ
rG/O . r2/G = rl2 cos φ.
The approximate moment about G is
MG ≈ −
3 GmO
m1l12 + m2l22 sin 2φ.
2 r3
We can simplify this result a bit further by recognizing that the quantity in parentheses is just the moment of inertia of the satellite about its center of mass: IG =
m1l12 + m2l22. The simplified expression is
MG ≈ −
3 GmO
IG sin 2φ.
2 r3
(9.73)
The relationship of the external gravitational moment to the moment of inertia turns
out to hold for any symmetric satellite. You can explore this relation in Problem 9.3.
Note that the moment in Eq. (9.73) depends on the satellite’s position coordinate r.
Thus the rotational equations of motion do not completely separate from the translational equations of motion; the external moment depends on the position in orbit.
However, recall that Example 4.9 considered a satellite in a circular orbit
of radius r0
and determined that the angular rate was constant and equal to θ̇0 = GmO /r03. In
that case, the moment on a two-mass satellite in a circular orbit reduces to
3
MG ≈ − θ̇02IG sin 2φ,
2
which is independent of the translational motion.
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CHAPTER NINE
To find the equation of motion for the satellite in a circular orbit, we first calculate
its angular momentum about the center of mass:
hG = IGIωB = IG
I
I C
ω + CωB = IG θ̇0 + φ̇ b3.
Using Euler’s second law in Eq. (9.19), we take the inertial derivative of IhG and set
it equal to the external moment about G to find the approximate equation of motion:
3
IGφ̈ = − θ̇02IG sin 2φ,
2
where we have used the fact that, for a circular orbit, θ̈ = 0. Dividing both sides by
IG gives the final equation of motion:
3
φ̈ + θ̇02 sin 2φ = 0.
2
(9.74)
Eq. (9.74) looks just like the equation of motion for a simple pendulum! In fact, if
we assume that the satellite starts aligned with er (the local vertical) and has only a
small deviation φ, the equation of motion can be approximated using the small-angle
formula:
φ̈ + 3θ̇02φ = 0.
This is the familiar equation for simple harmonic motion.
The dumbbell satellite
√
oscillates about the local vertical at a frequency of 3 times its orbit rate. This is
why satellites are often designed to be long and thin or are equipped with a long
boom and a mass at its end. If the goal is to keep it pointed toward the earth (e.g.,
to take scientific data or relay communications), the moment from the gravitational
gradient will keep it roughly pointed in that direction.
Tutorial 9.3 The Falling Chimney
The falling chimney is a classic problem in dynamics. Figure 9.22a is a photograph
of a famous industrial smokestack falling over in Glasgow, Scotland. Notice that it
breaks as it falls. Our goal is to find the distance from the bottom of the chimney
where it will break (if at all) and at what angle of fall.
We approach this problem by first solving for the equations of motion of the entire
chimney falling as a rigid body pinned at its base. This problem is an inversion of that
of the compound pendulum already solved but is otherwise identical. Figure 9.22b
and c show the coordinate θ and the reference frames used. We “break” the chimney
at some distance x from the bottom and treat each segment as a linked rigid body.
We include the constraint forces—which in this case are the internal shear stress,
normal stress, and moment in the chimney—and solve for the motion of each segment.
By using the constraint that, before the break, the two segments must be moving
together, we can solve for the internal forces. The chimney will break at the location
of maximum stress.
We begin by treating the entire chimney as an inverted compound pendulum. (Note
that we are ignoring the very slight error introduced because the chimney actually
tips on its corner rather than on an attachment point through the centerline.) As in
DYNAMICS OF A PLANAR RIGID BODY
391
θ
g
l
B
I
e2
b2
b1
O
(a)
(b)
O
e1
(c)
Figure 9.22 (a) A falling chimney. (b) Coordinates. (c) Reference frames.
Example 9.15, we sum moments about the attachment point O and set them equal to
the change in angular momentum:
IO θ̈b3 = MO = mg
l
sin θ,
2
where we used the following transformation array between body frame B and inertial
frame I :
e1
b1
b2
cos θ
− sin θ
e2
sin θ
cos θ .
As in the compound pendulum, the reaction forces at the base do not contribute to the
moment about O.
The moment of inertia of a long rod about an axis through its center of mass is
IG = ml 2/12 (see Appendix D). Using the parallel axis theorem (Eq. (9.34)), the
inertia about O is IO = ml 2/3. We can then solve for the angular acceleration of the
chimney:
θ̈ =
3g
sin θ.
2l
(9.75)
Just as for the simple pendulum, inverting the compound pendulum changes the sign
on the moment term, resulting in an unstable solution (which may be why it tips!).
Next we find the equations of motion for the upper segment of the chimney using
the free-body diagram shown in Figure 9.23b. We apply the separation principle and
solve for the motion of the center of mass and the motion about the center of mass. The
density of the chimney is assumed to be uniform, so the mass of the upper segment is
a linear fraction of the total mass: (1 − x/ l)m. The translational equations of motion
for the center of mass G of the upper segment are
x
x
1−
mIaG/O = V b1 + N b2 − 1 −
mg(sin θb1 + cos θb2), (9.76)
l
l
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CHAPTER NINE
G
– 冢1 – –x 冣mge2
l
Vb1
τb3
P
Nb2
–mge2
x
Fnb2
Ft b1
O
O
(a)
(b)
Figure 9.23 (a) Free-body diagram of the rigid rod. (b) Free-body diagram of the segment
above the cut.
where V is the shear force on the segment, and N is the axial normal force. The
rotational equation of motion about the center of mass due to the internal bending
moment 4 τ = τ b3 and the moment about G from the shear force is found using
Eq. (9.19):
l
x
IGθ̈ = τ + V
1−
,
(9.77)
2
l
where the moment of inertia IG of the segment about its center of mass is given by
IG =
1
12
1−
x
l
m(l − x)2 =
3
1 2
x
ml 1 −
.
12
l
We have also used the kinematic constraint that, before the chimney breaks, the
rotation angle of the upper segment must be the same as the rotation angle of the entire
chimney (modeled as a compound pendulum). In other words, the angular acceleration
θ̈ in Eq. (9.77) is the same as that in Eq. (9.75).
To solve for the shear and internal bending moment requires determining IaG/O
in Eq. (9.76). We find this quantity by using the kinematic constraint that the top
segment must travel with the rest of the rod (as they are a solid piece). Thus point P
at the top of the lower segment must have the same acceleration as the matching point
on the upper segment. We thus use the results of Chapter 8. Starting with the lower
segment, since P is fixed in the body frame attached to this segment and frame B is
not translating, the inertial acceleration of P is found from Eq. (8.21) to be
aP/O = IαB × rP/O + IωB ×
I
I B
ω × rP/O .
Using the position of P , rP/O = xb2, the acceleration of P is
I
aP/O = −x θ̈b1 − x θ̇ 2b2.
4 The
(9.78)
internal bending moment in a rod is an example of a pure torque. We thus use τ to represent it.
DYNAMICS OF A PLANAR RIGID BODY
393
Next we find the acceleration of P on the upper segment. To do that, we take the
derivative of the vector triad,
rP/O = rG/O + rP/G,
and use the fact that rP/G = − 21 (l − x)b2 to find
I
vP/O
l
= vG/O −
2
I
l
= vG/O +
2
I
x
1−
l
x
1−
l
Id
dt
b2
θ̇b1.
Taking the inertial derivative a second time gives
l
x
l
x
I
aP/O = IaG/O +
1−
θ̈b1 +
1−
θ̇ 2b2.
2
l
2
l
(9.79)
Setting Eqs. (9.78) and (9.79) equal to each other gives a kinematic expression for
the center-of-mass acceleration of the top segment:
l
x
I
aG/O = −
1+
θ̈b1 + θ̇ 2b2 .
2
l
We can now use this expression for the inertial acceleration of G in Eq. (9.76) to
find the shear force. From the b1 component we have
l
x
x
x
mθ̈.
mg sin θ = −
1+
1−
V − 1−
l
l
2
l
Substituting from the equation of motion for θ̈ in Eq. (9.75) gives an expression for
the shear force as a function of x and the angle of the chimney:
x
x
1
1−
1− 3
mg sin θ.
(9.80)
V =
4
l
l
Eqs. (9.76) and (9.77) and the shear force in Eq. (9.80) can be used to solve for
the internal bending moment of the chimney:
τ=
x
4
1−
x
l
2
mg sin θ.
(9.81)
Figure 9.24 plots the internal bending moment and shear force on the upper
segment from Eqs. (9.80) and (9.81). From Newton’s third law, the corresponding
moment and shear on the lower segment are equal and opposite. A falling chimney
is likely to fail in two possible ways, depending on the details of construction. As the
maximum shear occurs at the base (and is equal to the tangent reaction force), many
chimneys will fail by shearing off at the bottom. More likely, the chimney will fail
in bending. Figure 9.24 shows that the maximum bending moment occurs at exactly
394
CHAPTER NINE
Normalized shear and moment in chimney
3.0
12V/(mg sin θ)
27τ/(mgl sin θ)
2.5
2.0
–x = 1–
l 3
1.5
–x = 2–
l 3
1.0
0.5
0.0
–0.5
–1.0
0.0
0.2
0.4
0.6
0.8
1.0
–x
l
Figure 9.24 Normalized bending moment (dashed line) and shear force (solid line) on upper
segment as functions of distance along chimney. (Note: the moment has been scaled so that
the maximum occurs at 1 in the plot.)
1/3 the distance up from the base. The angle at which it fails depends on the strength
of the chimney, as the moment at x increases sinusoidally with θ. A fun experiment
is to build a tower of blocks and pull out the bottom; the tower should split apart at
exactly 1/3 of the total number of blocks (it helps to use a multiple of three blocks).
It is also interesting to note that the moment on the lower segment is negative
(because it is equal and opposite to that on the top segment). Thus a cross section of
the chimney is subject to tension at its lower edge and compression at its upper edge
as it falls. Since brick or cement chimneys are constructed to support compression,
they will always fail at the lower edge first, causing the top segment to fall slightly
backward, as seen in Figure 9.22a.
Finally, it is interesting to perform a somewhat more detailed analysis of the stress
in the chimney on a cross section. The stress depends on the moment, the axial force,
and the angle of fall. It turns out that, depending on the cross-section size, the length
of the chimney, and the angle of rupture, failure occurs somewhere between 1/3 and
1/2 the height of the chimney. That is why the failure in Figure 9.22a occurs closer
to 1/2 than 1/3. You will explore this phenomenon in more depth in Problem 9.7.
9.8 Key Ideas
.
.
A rigid body is a collection of particles rigidly attached to one another. That is,
each particle is constrained to remain motionless relative to every other particle.
The position of the center of mass of a continuous rigid body B with density
ρ(rdm/O ) is
DYNAMICS OF A PLANAR RIGID BODY
rG/O =
.
395
1
mG
rdm/O ρ(rdm/O )dV .
B
The center of mass of a rigid body obeys Euler’s first law,
FG = mGIaG/O .
.
The angular momentum of a rigid body about an inertially fixed point O obeys
Euler’s second law,
Id
I
hO = M O .
dt
Likewise, the angular momentum of a rigid body about its center of mass also
obeys Euler’s second law,
Id
I
hG = M G .
dt
.
The moment of inertia IG of a planar rigid body about its center of mass G is the
sum of the mass-weighted squared distance of each point on the body to the center
of mass. For a rigid collection of N particles,
N
IG =
mi ri/G2
i=1
and, for a continuous rigid body B,
IG =
.
B
rdm/G2dm.
The angular momentum with respect to inertial frame I of a planar rigid body
B about its center of mass G is
hG = IGIωB = IGIωB b3,
I
where B is a frame attached to B.
.
.
A pure torque τ is a moment acting on a rigid body that does not arise from a
specific, unique force acting at a known location.
The rotational dynamics of a planar rigid body with constant inertia are
Id
dt
hG = IGIα B b3 = MG.
I
For a set of discrete forces, the total moment is
MG =
N
i=1
ri/G × Fi(ext),
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CHAPTER NINE
where Fi(ext) is the total external force acting on particle i. For a continuous field
force acting on the body, we have
MG =
B
(ext)
dm,
rdm/G × fdm
where fdm is the force per unit mass on the body.
.
The dynamics of a rigid body undergoing an angular impulse are
I
hG(t2) = IhG(t1) + MG(t1, t2),
where
MP/O (t1, t2) =
t2
MP/O dt.
t1
.
The dynamics of a planar rigid body about an arbitrary body point Q are
Id
dt
hQ = IQIα B b3 = MQ + rQ/G × mGIaQ/O ,
I
where
MQ = MG − rQ/G ×
N
Fi(ext).
i=1
.
The parallel axis theorem relates the moment of inertia of a rigid body about an
arbitrary body point Q to the moment of inertia about the center of mass G:
IQ = IG + mGrQ/G2.
.
The radius of gyration kQ relative to a point Q of a rigid body is the equivalent
distance from Q of a particle of mass mG needed to produce the same moment of
inertia as the rigid body about Q:
2 = IQ .
m G kQ
.
The total kinetic energy of a rigid body both translating relative to an inertial
origin O and rotating about its center of mass G is
TO = TG/O + TG,
where the kinetic energy about the center of mass is
1
1
TG = IGIωB 2 = IhG . IωB .
2
2
.
The total work on a rigid body is
W = WG/O + WG = TO (t2) − TO (t1),
DYNAMICS OF A PLANAR RIGID BODY
397
where WG/O = WG/O (rG/O , θ; γG, γθ ) is the work due to the total force acting on
the rigid body along paths γG and γθ , and
t2
WG(rG/O , IωB ; γG, γθ ) =
MG . IωB dt
t1
is the work due to moments acting about the center of mass.
.
External work satisfies the conservation law for kinetic energy of the center of
mass and kinetic energy about the center of mass:
WG/O = TG/O
WG = TG.
.
The total energy of a rigid body B is
TG/O (t) + TG(t) + UO (t),
EO (t) =
where, for a set of discrete forces,
UO (t) =
N
Ui/O (ri/O (t)),
i=1
and, for a continuous field force acting on the body,
UO (t) =
.
B
Udm/O (rdm/O )dm.
The work-energy formula for a rigid body relates the change in total energy to
the nonconservative work:
(nc)
EO (t2) = EO (t1) + WG/O
+ WG(nc).
9.9 Notes and Further Reading
We have taken a decidedly Newtonian approach in our development in this book. We
have done so primarily for pedagogical reasons, as the material is more approachable
and easier to learn by starting with particle mechanics, moving to multiparticle
systems, and then on to rigid bodies. We have thus followed most contemporary
physics and engineering texts in treating a rigid body as a collection of particles
and then deriving Euler’s laws. As pointed out in the text, this approach requires
making an additional axiomatic assumption about internal moments of a rigid body
that does not follow from Newton’s third law. This difficulty caused Euler to abandon
the particle or “corpuscular” model for rigid bodies entirely and postulate his laws
of motion as axiomatic for rigid bodies, just as Newton’s laws are for particles. This
approach eliminates the need for any assumptions about the internal forces in a rigid
body and makes the development of continuum mechanics more natural. It also has
some logical appeal, since all bodies studied in dynamics are in fact rigid; particles
are merely an abstraction. Thus Euler’s laws can be considered more fundamental and
general than Newton’s. An excellent discussion of the history of Euler’s treatment of
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CHAPTER NINE
rigid bodies and the nature of Euler’s laws can be found in Truesdell (1968). Rao
(2006) is an example of a recent text that presents Euler’s laws as axiomatic rather
than derivable from Newton’s.
Only a few other texts make the distinction between moments and pure torques.
Our notation came from Rao (2006) and Tenenbaum (2004). The dynamics of pool is
a fascinating and complex subject, incorporating collisions, rigid-body dynamics, and
complex models of friction. Fetter and Walecka (1980) has an insightful discussion
(our pool ball example came from there). The yo-yo de-spin problem is a classic example in spacecraft attitude dynamics and control. It appears in many texts, including
Kaplan (1976) and Wertz (1978). Our example closely follows that in Kaplan (1976).
The falling chimney problem is also a classic problem in introductory courses, combining topics from both statics and dynamics. Variations can be found in many texts,
including Bedford and Fowler (2002). There is a particularly interesting short paper
by Varieschi and Kamiya (2003) that we used as a reference for both the tutorial and
problem.
9.10 Problems
9.1
Find the position of the center of mass rG/O of the planar rigid body in
Figure 9.25. The rigid body has uniform density, with m = 2 kg, r1 = 0.3 m,
and r2 = 0.4 m.
B
r2
r1
O
Figure 9.25 Problem 9.1.
9.2
Find the moment of inertia IG of the planar rigid body in Figure 9.26. The
rigid body has uniform density, with m = 2 kg, w = 0.3 m, and h = 0.6 m.
B
h
G
w
Figure 9.26 Problem 9.2.
DYNAMICS OF A PLANAR RIGID BODY
399
9.3
Show, using the moment definition in Eq. (9.12), that the gravitational gradient
moment on any long, axisymmetric satellite is given by Eq. (9.73).
9.4
Following Example 9.11, suppose the cue ball is hit dead center (h = 0). If
the coefficient of sliding friction between the cue ball and the felt is μ, how
long will the cue ball slide before it starts rolling without slipping?
9.5
Recall the sliding collar in Problem 3.20 in Chapter 3. It is shown again in
Figure 9.27 with an added twist. Rather than being fixed, the shaft, with
dimensions shown and mass M, is connected to the pivot point O by two
massless rods and is allowed to swing in the plane like a pendulum. The collar
has mass m (and can be treated as a point mass) and the shaft is a distance l
away from O. The spring has spring constant k and its unstretched length is l.
The position of the mass along the shaft is x and the angle of the perpendicular
to the shaft with the local vertical is θ.
θ
O
l
M
k
g
x
L
m
b
Figure 9.27 Problem 9.5.
a. How many degrees of freedom are there?
b. Show that the total angular momentum of the system is
M(L2 + b2)
2
2
I
hO =
+ (M + m)l + mx θ̇ez − ml ẋez .
12
Is IhO conserved?
c. Show that the equations of motion for the system are
M(L2 + b2)
2
2
+ Ml + mx θ̈ + 2mx ẋ θ̇ − mlx θ̇ 2 + Mgl sin θ
12
l
−mgx cos θ + kxl 1 − √
=0
x2 + l2
k
l
2
− g sin θ = 0.
ẍ − l θ̈ − x θ̇ + x 1 − √
m
x2 + l2
d. Is total energy conserved?
400
CHAPTER NINE
e. Suppose we start the system with the shaft at a positive angle of
θ (0) = 30◦ and the point mass/collar at a positive stretch of x(0) = 60
cm. Setting L = 2 m, l = 2 m, b = 10 cm, m = 0.25 kg, M = 10 kg,
k = 1 N/m, and g = 9.8 m/s2, use matlab to simulate the motion of
the shaft and the collar over the time interval [0, 30] s. Plot x versus t,
θ versus t, and IhO versus t.
9.6
Consider the rotating disk with an offset slot, as shown in Figure 9.28. Inside
the slot is a particle P of mass m connected to a spring with spring constant
k. The disk has mass M and moment of inertia I about the center axis (not
including the small mass m). The disk is free to rotate about its axis by an
angle θ. You can assume that the unstretched length of the spring is zero and
at t = 0 the spring is stretched by an amount x0.
x
P
l
θ
O
r
Figure 9.28 Problem 9.6.
a. Show that the equations of motion for the mass in the slot and the angle
θ of the disk are
θ̈ =
−2mx ẋ θ̇ + mlx θ̇ 2 − klx
I + mx 2
ẍ = l θ̈ + x θ̇ 2 − (k/m)x.
b. Is energy conserved?
c. What is the total angular momentum IhO of the system? Is it conserved?
d. Suppose we start the system by offsetting the mass an amount x0 = 60
cm. If the disk has a radius of 1 m, I = 2.5 kg-m2, l = 75 cm, m = 0.25
kg, and k = 1 N/m, use matlab to simulate the motion of the mass and
the disk over the time interval [0, 20] s. Plot x versus t, θ versus t, and
IhO versus t.
9.7
In this problem you examine a slightly more realistic model for breakage of
the falling chimney in Tutorial 9.3. You may recall from a statics class that
the stress in a cross section can be written as a function of the axial force and
DYNAMICS OF A PLANAR RIGID BODY
401
bending moment. In particular, the maximum stress at the edges of the cross
section (in tension and compression) is
σ=
aτ
N
∓
,
A
2J
where A is the cross sectional area, a is the side length of a square cross
section, and J is the area moment of inertia, J = a 2/12. Our calculation of
where the chimney will break in Tutorial 9.3 was a bit simplified. Rather than
failing at the maximum shear or bending moment point, it will in fact fail at
the point of maximum internal stress (in particular, tension on the lower edge).
a. As we did for shear and moment in Tutorial 9.3, find an expression for
the axial force in the chimney as a function of the position x along the
chimney.
b. Find an expression for the internal stress of the chimney as a function
of position x along the chimney, angle of fall θ, and aspect ratio l/a.
c. Plot the stress in the chimney versus distance along the chimney for
various fall angles and aspect ratios. Show that, depending on the
aspect ratio (tall thin chimneys versus short fat ones), it will fail either
at 1/3 or 1/2 the distance along the chimney.
9.8
Using the results of Problem 9.2, consider the compound crane illustrated in
Figure 9.29. A compound pendulum is attached to mass M at point Q. Mass
M = 2 kg slides horizontally without friction. The distance between Q and G
is 0.25 m.
a. Find the equations of motion in x and θ.
b. Integrate the equations of motion in matlab, and create a matlab file
that animates the solution.
I
M
g
x
Q
O
G
θ
Figure 9.29 Problem 9.8.
9.9
Recall the simple double pendulum explored in Example 8.9. Repeat the
derivation of the equations of motion for the double pendulum, but now model
it as two connected rigid bodies with moments of inertia (about their centers
of mass) I1 and I2, masses m1 and m2, and lengths l1 and l2. Solve for the
equations of motion, assuming the links are both thin rods.
9.10 A thin rigid cylindrical rod is leaned against a wall as in Figure 9.30, with its
other end resting on the floor. The rod slides (without friction) under the force
402
CHAPTER NINE
l
θ
Figure 9.30 Problem 9.10.
of gravity. Find the equations of motion for the angle θ that the rod makes
with the wall.
9.11 For the sliding rod in Problem 9.10, find the angle θ at which the rod loses
contact with the wall, assuming the rod starts sliding with θ(0) = 0 and θ̇(0)
small.
9.12 In Figure 9.31 a flatbed truck of mass M, containing a crane modeled as a
long thin rod pivoted at its end, suddenly accelerates forward. Assume the rod
is 2 m long, weighs 50 kg, and leans at an angle θ of 60◦ from the flatbed.
B
A
a
θ
Figure 9.31 Problem 9.12.
a. If the acceleration is 5 m/s2, what is the normal force on the crane rod
at B?
b. What is the largest possible acceleration a before the rod will lose
contact with the truck at B?
9.13 Consider a wheel of radius R2, illustrated in Figure 9.32. String is wrapped
around a concentric spool of radius R1, making an angle of α with the horizonR2
g
T
R1
α
Figure 9.32 Problem 9.13.
DYNAMICS OF A PLANAR RIGID BODY
403
tal. When tension T is applied to the string, the wheel rolls without slipping
in a direction (left or right) determined by the angle α. Find the critical angle
α ∗ above which the wheel rolls to the left and below which it rolls to the right.
9.14 A uniformly dense marble of mass m = 0.05 kg and radius R = 0.01 m is
released from rest at the top of a ramp and rolls without slipping down the
ramp and off a table, as shown in Figure 9.33. Find the distance d from the foot
of the table where the marble lands on the floor. The ramp height is h1 = 0.2
m, and its width is w1 = 0.4 m. The marble rolls a distance of w2 = 0.15 m
on the table, which has height h2 = 1 m. [HINT: Consider both the rotational
and translational motion of the rolling marble.]
g
m
h1
w2
w1
h2
d
Figure 9.33 Problem 9.14.
9.15 A marble rolls without slipping on a walkway that accelerates horizontally
with constant acceleration a relative to an inertially fixed point O, as shown
in Figure 9.34. Find, in terms of a, the acceleration of the marble’s center
of mass G relative to O. For the moment of inertia of the marble, assume
IG = (2/5)mR 2, where m and R are the mass and radius of the marble,
respectively. [HINT: Consider both the translational and rotational dynamics
of the marble.]
I
s
G
R
θ
a
O
Figure 9.34 Problem 9.15.
9.16 A wheelchair of mass m rolls without slipping up a ramp of angle φ, as shown
in Figure 9.35. Find the horizontal force T necessary to roll the wheelchair up
the ramp at a constant speed. Assume that force acts at a distance r from the
center of the wheel, which has radius R. Also assume that the center of mass
G of the wheelchair is located in line with the center of the wheels.
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CHAPTER NINE
g
R
T
r
G
ϕ
O
Figure 9.35 Problem 9.16.
9.17 Returning to the particle on a beam system of Example 9.21, replace the
particle with a flat disk that rolls on the beam without slipping.
a. How many degrees of freedom are there in this modified system?
b. Find the equations of motion of the disk and beam.
9.18 Sketch a model of two trapeze artists on a trapeze using two coupled rigid rods
(see Figure 9.36). How many degrees of freedom does this system have? Select
coordinates and find the equations of motion for each artist. [HINT: Consider
a double-pendulum model.] Choosing reasonable values for the parameters in
your model, integrate the equations of motion in matlab and plot the solution
for several different initial conditions. Be sure to label your axes and add a
legend to each plot.
Figure 9.36 Problem 9.18.
9.19 A uniform disk rotor with moment of inertia IG, similar to the flywheel in
Figure 9.16, is connected to a DC motor with input voltage V . The current i
in the armature circuit is governed by Kirchhoff’s current law and can be
expressed as
L
di
+ Ri + k θ̇ − V = 0,
dt
DYNAMICS OF A PLANAR RIGID BODY
405
where L is the electric inductance, R the resistance, k the electromotive force
constant, and θ̇ is the rotation rate of the rotor. The torque applied to the rotor
by the motor is
τ = ki,
and the mechanical linkage between the motor and rotor has a damping ratio b.
a. Show that the equation governing the motion of the rotor is
IGθ̈ = −bθ̇ − ki.
b. Using matlab, integrate the two governing equations of this system.
Set the rotor moment of inertia IG to 0.01 kg-m2, the damping ratio b
to 0.2 N-m-s, the resistance R to 1 ohm, the inductance h to 0.75 H,
and the electromotive force constant k to 0.03 N-m/amp. Plot the rotor
torque as a function of time for 5 s after a constant voltage of 1 V is
applied to a previously turned off motor (i.e., all initial conditions are
zero).
c. Repeat the previous part, but now plot the torque as a function of time
when the voltage is ramped from 0 to 1 V over a 5 s period (integrate
for at least 10 s).
9.20 Two identical particles of mass m are suspended on a rigid rod of length l to
form a dynamic pendulum, as shown in Figure 9.37. Particle P is fixed to one
end of the rod; the other end is attached to a fixed pivot. Particle Q, which is
free to slide up and down the rod, is connected to P by a spring of constant k
and rest length l0.
O
Q
g
l
k
P
Figure 9.37 Problem 9.20.
a.
b.
c.
d.
Find the internal potential energy UO(int) of the system.
Find the external potential energy UO(ext) of the system.
(tot)
Find the total energy EO
of the system.
Is the total energy conserved?
9.21 The Cub Scouts has an annual competition called the Pinewood Derby. Each
scout builds a wooden car that will race down an inclined track (inclined at
some angle θ) powered only by gravity. The goal is to build a car that will
make it down the slope as fast as possible. The total mass of the car (including
the wheels) must be ≤ M, and the radius of the wheels of the car must be
R, though the wheel weight can be any fraction of the total you like. In this
406
CHAPTER NINE
problem you will investigate whether it is better to make the wheels heavy or
light (with ballast added to the car body).
Derive an expression for the time it takes the car to reach a certain position
x down the incline in terms of R, θ, g, M, and the mass mw of one of the
four wheels. From this expression, make a prediction as to which type of car
(one with heavier or lighter wheels) will get to the bottom of the track fastest.
Ignore air resistance and assume that each of the four wheels rolls without
slipping while going down the track.
9.22 A thin uniform rod of mass mr and length l is fitted with a small collar of mass
mc that can slide freely (without friction), as shown in Figure 9.38. A motor at
the rod’s attachment point O rotates the rod with a constant torque τ. Assume
the system is acting in plane (so that gravity can be neglected).
mr
l
mc
θ
O
Figure 9.38 Problem 9.22.
a. Find the equations of motion of the collar.
b. Using matlab, integrate the equations of motion found in part (a).
Let τ = 1 N-m, l = 1 m, mc = 1 kg, and mr = 1/3 kg. Start the system
from rest with θ = 0 and with the collar 0.1 m from O and integrate
for 1 s. Plot the collar’s position along the length of rod and the rod’s
angle θ versus time.
c. Integrate the system in part (b) until the collar leaves the rod and find
the value of θ and the collar’s velocity at this time.
d. Repeat your integrations for at least two other values of τ between 1
and 3 N-m, and find the value of θ at which the collar leaves the rod
in each instance. Explain your findings.
PART FOUR
Dynamics in Three Dimensions
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CHAPTER TEN
Particle Kinematics and Kinetics in Three Dimensions
In this part we reach the final stage of our discussion of Newtonian dynamics. At last
we study the full three-dimensional motion of particles and rigid bodies. This chapter
reconsiders the kinematics of a particle, allowing motion in all three dimensions.
While many of our two-dimensional results apply with only minor modifications
or some new understanding, we will need to develop new and more sophisticated
approaches to describing orientation in three dimensions. The next chapter undertakes
the more challenging task of generalizing the motion of rigid bodies to rotation in three
dimensions.
10.1 Two New Coordinate Systems
Chapter 3 introduced the concepts of vectors, coordinate systems, and reference
frames and described how to find the velocity and acceleration of a particle in terms
of different coordinate systems and in different frames. Chapter 8 introduced the
study of the relative motion of a particle in a translating and rotating frame. Both
chapters treated motion in only two dimensions. We now relax that restriction. This
chapter repeats our entire treatment of particle dynamics, allowing motion in all three
dimensions. We start by introducing two new coordinate systems—cylindrical and
spherical coordinates1—that can be used to describe the configuration of a particle.
But first we reprise our fundamental coordinate system—Cartesian coordinates—to
set the framework for the three-dimensional analysis.
10.1.1 Cartesian Coordinates in Three Dimensions
Recall that we can locate a particle in an inertial reference frame most directly by
using Cartesian coordinates (x, y, z)I :
rP/O = xex + yey + zez ,
1 We
do not address three-dimensional path coordinates, as this is a more advanced subject.
410
CHAPTER TEN
I
I
ez
ez
P
P
rP/O
rP/O
z
ϕ
r
ey
ey
O
O
r
ex
θ
ex
θ
(b)
(a)
Figure 10.1 (a) Cylindrical (r, θ, z)I and (b) spherical (r, θ, φ)I coordinates locate point P
in reference frame I in three dimensions.
where I = (O, ex , ey , ez ). The inertial velocity and acceleration of P are found from
vector differentiation and Definition 3.3:
I
Id
I
Id
vP/O =
aP/O =
dt
dt
rP/O = ẋex + ẏey + żez
vP/O = ẍex + ÿey + z̈ez .
I
Finding the equations of motion in Cartesian coordinates involves applying Newton’s
second law to each of the three directions, just as in Chapter 3.
What about other coordinate systems? As in the planar case, it is often convenient
to use another coordinate system to describe the position of P in I (usually because
of constraints in the problem). The two most common three-dimensional coordinate
systems other than Cartesian coordinates are the cylindrical coordinates shown in
Figure 10.1a and the spherical coordinates shown in Figure 10.1b.
10.1.2 Cylindrical Coordinates
Cylindrical coordinates (r, θ, z)I are a simple generalization of polar coordinates.
All previous results for polar coordinates apply to the components of motion in the
plane defined by ex and ey ; we simply add the third Cartesian coordinate z in the ez
direction and apply Newton’s second law in that direction. Thus the position of P
with respect to O in cylindrical coordinates is
rP/O = r cos θex + r sin θey + zez
(10.1)
and the inertial kinematics expressed as components in I are
I
vP/O = (ṙ cos θ − r θ̇ sin θ)ex + (ṙ sin θ + r θ̇ cos θ)ey + żez
(10.2)
I
aP/O = (r̈ cos θ − 2ṙ θ̇ sin θ − r θ̈ sin θ − r θ̇ 2 cos θ)ex
+ (r̈ sin θ + 2ṙ θ̇ cos θ + r θ̈ cos θ − r θ̇ 2 sin θ)ey + z̈ez .
(10.3)
Compare these equations to Eqs. (3.18) and (3.19). They are identical except for the
addition of the rates in the z-direction; we thus solve dynamics problems in cylindrical
PARTICLE KINEMATICS AND KINETICS IN THREE DIMENSIONS
411
coordinates just as in polar coordinates and all previous results apply. The complexity
of these equations, however, leads us again to ask whether a different frame would
further simplify the kinematics. The answer is yes, but we defer that discussion until
after introducing spherical coordinates.
10.1.3 Spherical Coordinates
The spherical coordinates (r, θ, φ)I illustrated in Figure 10.1b are the threedimensional analogue of polar coordinates. They consist of the radial distance r
of point P from O, the azimuth angle θ, and the polar angle φ. Simple geometric
considerations give the components of the vector rP/O expressed in terms of spherical
coordinates:
rP/O = r cos θ sin φex + r sin θ sin φey + r cos φez .
(10.4)
Just as for Cartesian and cylindrical coordinates, the velocity and acceleration of
P relative to I can be found by differentiating Eq. (10.4) and using the definition of
the vector derivative and the chain rule:
I
vP/O = (ṙ sin φ cos θ + r φ̇ cos φ cos θ − r θ̇ sin φ sin θ)ex
+ (ṙ sin φ sin θ + r φ̇ cos φ sin θ + r θ̇ sin φ cos θ)ey
+ (ṙ cos φ − r φ̇ sin φ)ez
(10.5)
I
aP/O = (r̈ sin φ cos θ + r φ̈ cos φ cos θ − r θ̈ sin φ sin θ + 2ṙ φ̇ cos φ cos θ
− 2ṙ θ̇ sin φ sin θ − r φ̇ 2 sin φ cos θ − r θ̇ 2 sin φ cos θ − 2r φ̇ θ̇ cos φ sin θ)ex
+ (r̈ sin φ sin θ + r φ̈ cos φ sin θ + r θ̈ sin φ cos θ + 2ṙ φ̇ cos φ sin θ
+ 2ṙ θ̇ sin φ cos θ − r φ̇ 2 sin φ sin θ − r θ̇ 2 sin φ sin θ + 2r φ̇ θ̇ cos φ cos θ)ey
+ (r̈ cos φ − r φ̈ sin φ − 2ṙ φ̇ sin φ − r φ̇ 2 cos φ)ez .
(10.6)
These equations can be used in Newton’s second law to find equations of motion
for the three spherical coordinates of a particle. We rarely do so, however, since
these expressions are even more complicated and algebraically messy than their polarcoordinate counterparts. This strongly motivates us, again, to examine whether the
introduction of a new reference frame would simplify things. We do so after a couple
of simple examples.
Example 10.1 The Spherical Pendulum
Chapters 3 and 4 introduced the concepts of coordinate systems, vector derivatives,
and equations of motion by examining the simple pendulum with varying levels
of sophistication. We do the same here using the spherical pendulum, shown in
Figure 10.2. The pendulum mass is a distance l away from the attachment point,
just as in the simple pendulum, only here it is free to swing in two directions. It is
thus a two-degree-of-freedom problem, which requires two coordinates to describe
the configuration.
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CHAPTER TEN
I
ϕ
ez
O
g
ex
ey
l
β
F
mP
θ
(a) Coordinates
–mPgez
(b) Reference frame
(c) Free-body diagram
Figure 10.2 Spherical pendulum using spherical coordinates (r, θ, φ)I .
The geometry of the pendulum is such that it is naturally described in terms of the
spherical coordinates (r = l, θ, φ = π − β)I , as in Figure 10.2a. It is thus sensible
to find equations of motion for the two angles θ and φ. We could do so by combining
the free-body diagram with the acceleration from Eq. (10.6), as done in Chapter 3
for the simple pendulum. However, as stated above, this is rarely done because of the
complexity of isolating the two coordinates. We wait, therefore, until after we have
introduced some three-dimensional rotating frames.
Example 10.2 Latitude and Longitude
A good example of applying spherical coordinates is the use of latitude and longitude
to locate a point on the surface of the earth. Figure 10.3 shows a point P located
at coordinates (λ, L)G on a spherical earth. The longitude λ corresponds to the
spherical angle θ. The latitude L is the complement of the spherical angle φ, so that
L = π/2 − φ (φ is sometimes referred to as the co-latitude).
The frame G = (O, g1, g2, g3) in which these spherical coordinates are defined
is located at the center of the earth O and fixed to the rotating surface. G is called
a geographic reference frame. This frame is defined by directions on the earth that
provide the zero reference for latitude and longitude. By international agreement, g3
is along the spin-axis direction and g1 is in the equatorial plane and directed toward
the zero meridian, which is the great circle that passes through Greenwich, England.
The unit vector g2 completes the right-handed set.
Because (λ, L)G are not exactly spherical coordinates, it is worthwhile to use
Eq. (10.4) to find the position of P with respect to O in terms of the latitude and
longitude. Using R for the radius of the earth, we have
rP/O = R cos λ cos Lg1 + R sin λ cos Lg2 + R sin Lg3.
Of course, the geographic frame is not an inertial frame because the earth is rotating
in absolute space. It has a simple angular velocity with respect to inertial frame I ,
also located at O and fixed to the distant stars, which implies
I G
ω =
E g3.
PARTICLE KINEMATICS AND KINETICS IN THREE DIMENSIONS
G
413
g3
ΩE
P
rP/O
L
O
λ
g2
g1
Figure 10.3 Spherical earth with latitude and longitude locating point P on the surface.
The angular velocity IωG of G with respect to I is directed along an axis through the
north pole. We call the inertial frame centered at O the geocentric inertial frame. The
fact that G is not inertial has interesting consequences when trying to solve for the
equations of motion of particles traveling with respect to it (e.g., airplanes or ships).
Tutorial 10.1 explores these consequences.
10.2 The Cylindrical and Spherical Reference Frames
This section introduces two reference frames that are three-dimensional equivalents
of the polar frame in two dimensions. These frames rotate in I to stay aligned with
a particle and result in compact and convenient vector expressions for the position,
velocity, and acceleration, just as with the polar frame in two dimensions. Look back
over Section 3.4. We had two objectives there: (a) to write the velocity and acceleration
as components in a new, more convenient frame and (b) to use that frame to find
a streamlined approach to calculating the velocity and acceleration via the angular
velocity. The same motivations hold here. We just need to be much more careful in
treating the angular velocity.
10.2.1 The Cylindrical Frame
Cylindrical frame Bc = (O, er , eθ , ez ) (Figure 10.4a) shares much in common with
the polar frame. Here, however, the unit vector er does not always point at particle
P but rather at the projection of the position of P onto the plane spanned by ex and
ey . Thus the unit vectors er and eθ coincide with the polar-frame unit vectors. The ez
unit vector coincides with the corresponding inertial frame unit vector.
Just as in the planar discussion, we can write down a transformation table that
allows us to write the unit vectors of one frame in terms of the unit vectors of another
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CHAPTER TEN
I
I
ez
ez
er
Bs
P
rP/O
z
θ
ϕ
ey
eθ
O
P
rP/O
r
eθ
O
ey
θ
r
ex
ex
Bc
er
eϕ
(a)
(b)
Figure 10.4 Cylindrical and spherical reference frames, Bc = (O, er , eθ , ez ) and Bs =
(O, eφ , eθ , er ). (a) Cylindrical frame. (b) Spherical frame.
frame. Since the ez unit vectors of the two frames coincide, the three-dimensional
generalization of the transformation table is
er
eθ
ez
ex
ey
cos θ
− sin θ
0
sin θ
cos θ
0
ez
0
0
1 .
(10.7)
This is a three-dimensional version of our vector transformation table, in which one
unit-vector direction of the two frames coincides. Every transformation table that
results from a simple rotation about a single axis resembles Eq. (10.7); that is, it looks
like a two-dimensional table with a new row and a new column containing zeros and
one.
Eq. (10.7) can be used to express the kinematics of P in Eqs. (10.1)–(10.3) as
components in the cylindrical frame rather than the inertial one. Using Eqs. (10.1)–
(10.3) and Eq. (10.7), we obtain the following expressions:
rP/O = rer + zez
(10.8)
I
vP/O = ṙer + r θ̇ eθ + żez
(10.9)
I
(10.10)
aP/O = (r̈ − r θ̇ 2)er + (2ṙ θ̇ + r θ̈)eθ + z̈ez .
These expressions should come as no surprise, given the relationship between
cylindrical and polar coordinates. Note that, as in Section 3.4.2, Eqs. (10.8)–(10.10)
could also have been derived using IωBc = θ̇ez and the transport equation (Definition 8.1) to differentiate rP/O and IvP/O .
10.2.2 The Spherical Frame
The spherical frame Bs = (O, eφ , eθ , er ) is shown in Figure 10.4b. It consists of a
unit vector er directed along rP/O toward P , a unit vector eφ orthogonal to er and in
the plane defined by er and ez , and a unit vector eθ = er × eφ completing the right-
PARTICLE KINEMATICS AND KINETICS IN THREE DIMENSIONS
I
I
ez, a3
415
ez, a3
ϕ
Bs
er
A
A
a2
a 2 , eθ
ey
ey
θ
a1
a1
ex
ex
(a)
eϕ
(b)
Figure 10.5 Relating spherical frame Bs = (O, eφ , eθ , er ) to inertial frame I = (O, ex , ey , ez )
by means of two rotations by θ and φ, respectively, and intermediate frame A = (O, a1, a2 , a3).
(a) θ rotation about ex . (b) φ rotation about a2 .
handed set (which implies that eφ = eθ × er and er = eφ × eθ ). How do we describe
the orientation of Bs in I and relate the unit vectors defining Bs to those defining I ?
None of the unit vectors are aligned anymore, so it is no longer as simple as just adding
a new row and column to the two-dimensional transformation table. The answer is to
use an intermediate frame.
Consider a frame A initially aligned with I that is rotated about the ez axis by
angle θ, as shown in Figure 10.5a. This frame is identical to the cylindrical frame in
Figure 10.4a. The relationship between I = (O, ex , ey , ez ) and A = (O, a1, a2, a3)
is the transformation table for the cylindrical frame:
a1
a2
a3
ex
ey
cos θ
− sin θ
0
sin θ
cos θ
0
ez
0
0
1 .
(10.11)
We can also describe the relative orientation between spherical frame Bs =
(O, eφ , eθ , er ) in terms of the A unit vectors because it differs from A by only a
simple rotation of φ about the a2 = eθ axis, as shown in Figure 10.5b. This transformation table is
eφ
eθ
er
a1
a2
a3
cos φ
0
sin φ
0
1
0
− sin φ
0
cos φ .
(10.12)
Successively using the two transformation tables in Eqs. (10.11) and (10.12) allows
us to change between inertial frame I and spherical frame Bs . For example, we can
transform the position of P from components in Bs to components in I . Written as
components in the spherical frame Bs , the position of P is
rP/O = rer .
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Using the transformation table in Eq. (10.12), rP/O is written as components in A:
rP/O = r sin φa1 + r cos φa3.
We then use the second transformation table in Eq. (10.11) to write rP/O as components in I :
rP/O = r cos θ sin φex + r sin θ sin φey + r cos φez ,
which is the same as the spherical-coordinate position in Eq. (10.4).
We can also use these transformation tables successively on the unit vectors to
form a single array for transforming unit vectors between I and Bs . Writing out the
relationship between the unit vectors of I and A gives
ex = cos θa1 − sin θa2
ey = sin θa1 + cos θa2
ez = a3.
Then, using the transformation table in Eq. (10.12), we have
ex = cos θ cos φeφ − sin θeθ + cos θ sin φer
ey = sin θ cos φeφ + cos θeθ + sin θ sin φer
ez = − sin φeφ + cos φer ,
which can be written as a single transformation table
eφ
eθ
er
ex
ey
ez
cos θ cos φ
− sin θ
cos θ sin φ
sin θ cos φ
cos θ
sin θ sin φ
− sin φ
0
cos φ .
(10.13)
Eq. (10.13) is the three-dimensional version of the relative-orientation transformation
between the inertial frame and the spherical frame. Remember, the trick was to use
an intermediate frame to derive this transformation.
Eq. (10.13) can now be used to change Eqs. (10.5) and (10.6), expressed as
components in the inertial frame, to ones expressed as components in the spherical
PARTICLE KINEMATICS AND KINETICS IN THREE DIMENSIONS
417
frame, analogous to what we did in Chapter 3 for the polar frame. Making the proper
substitutions and performing a bit of algebra (actually, a lot) results in
I
vP/O = r φ̇eφ + r θ̇ sin φeθ + ṙer
(10.14)
I
aP/O = (2ṙ φ̇ + r φ̈ − r θ̇ 2 cos φ sin φ)eφ
+ (2ṙ θ̇ sin φ + 2r θ̇ φ̇ cos φ + r θ̈ sin φ)eθ
+ (r̈ − r φ̇ 2 − r θ̇ 2 sin2 φ)er .
(10.15)
Just as with the polar and cylindrical frames, the spherical frame expressions are
simpler and easier to work with than their inertial frame counterparts. Now it is much
easier to find equations of motion in terms of spherical coordinates.
Example 10.3 The Spherical Pendulum, Take 2
Example 10.1 introduced the spherical pendulum—the three-dimensional version
of the simple pendulum studied in Chapter 3. However, we postponed finding its
equations of motion (in terms of the spherical coordinates θ and φ) until after we
introduced the spherical frame. Let us find them now. For reference, Figure 10.6 again
shows the geometry, with the addition of the spherical frame Bs and the reaction force
in terms of the er unit vector in the free-body diagram.
The inertial kinematics of the pendulum mass relative to the inertial frame and
expressed in spherical coordinates in the spherical frame are given by Eqs. (10.14)
I
ez
eθ
ϕ
Bs
O
g
l
ex
er
ey
eϕ
β
–Fer
mP
θ
(a) Coordinates
–mPgez
(b) Reference frames
(c) Free-body diagram
Figure 10.6 Spherical pendulum using spherical coordinates (r, θ, φ)I and spherical frame
Bs = (O, eφ , eθ , er ). Also shown is the angle from the vertical, β = π − φ.
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CHAPTER TEN
and (10.15) with r = l and ṙ = r̈ = 0. The kinematics of P are
rP/O = ler
I
vP/O = l φ̇eφ + l θ̇ sin φeθ
I
aP/O = (l φ̈ − l θ̇ 2 cos φ sin φ)eφ + (2l θ̇ φ̇ cos φ + l θ̈ sin φ)eθ
− (l φ̇ 2 + l θ̇ 2 sin2 φ)er .
The equations of motion for the pendulum bob are found from Newton’s second
law by equating the forces from the free-body diagram in Figure 10.6c with the
acceleration times the mass, which yields
−F er − mP gez = mP (l φ̈ − l θ̇ 2 cos φ sin φ)eφ + mP (2l θ̇ φ̇ cos φ + l θ̈ sin φ)eθ
− mP (l φ̇ 2 + l θ̇ 2 sin2 φ)er .
(10.16)
Since it is easiest to solve Eq. (10.16) in the spherical frame, we write ez in terms of
eφ and er using Eq. (10.13):
−F er − mP g(− sin φeφ + cos φer ) = mP (l φ̈ − l θ̇ 2 cos φ sin φ)eφ
+ mP (2l θ̇ φ̇ cos φ + l θ̈ sin φ)eθ
− mP (l φ̇ 2 + l θ̇ 2 sin2 φ)er .
(10.17)
Eq. (10.17) corresponds to the three scalar equations in the er , eθ , and eφ directions,
−F − mP g cos φ = −mP l φ̇ 2 − mP l θ̇ 2 sin2 φ
(10.18)
0 = 2mP l θ̇ φ̇ cos φ + mP l θ̈ sin φ
(10.19)
mP g sin φ = mP l φ̈ − mP l θ̇ 2 cos φ sin φ.
(10.20)
Eqs. (10.19) and (10.20) provide the two equations of motion,
φ̈ − θ̇ 2 sin φ cos φ −
g
sin φ = 0
l
θ̈ + 2θ̇ φ̇ cot φ = 0,
(10.21)
(10.22)
while Eq. (10.18) provides an expression for the reaction force in the rod.
At first these may seem a bit surprising, since the sign on the gravity term seems
to have switched from the simple pendulum. However, this is only because we have
defined φ as the angle down from the vertical ez axis for spherical coordinates. If we
write the equations of motion in terms of the angle β = π − φ (see Figure 10.6a),
then the equations of motion become
β̈ + θ̇ 2 sin β cos β +
g
sin β = 0
l
θ̈ + 2θ̇ β̇ cot β = 0.
PARTICLE KINEMATICS AND KINETICS IN THREE DIMENSIONS
419
Spherical coordinates allowed for a straightforward derivation of the equations,
as well as lent insight to the problem. Additionally, we can use this formulation to
study equilibrium points and steady-state solutions of the spherical pendulum. As
with the planar pendulum, there are two equilibrium solutions, a stable one where the
pendulum hangs down vertically (β = 0) and an unstable one where the pendulum
is suspended upright (β = π). Unlike the planar pendulum, however, there is a third
steady-state solution where the pendulum hangs at a fixed angle while rotating in a
circle. You are invited to explore this phenomenon in Problem 10.8.
Although Eqs. (10.14) and (10.15) are simpler than Eqs. (10.5) and (10.6), it
took quite a bit of algebra (much of which we spared you) to get there. Can we use
the angular velocity to find an easier route to the kinematics, as we did for polar
coordinates? The answer is yes, and the process is almost identical, but the concept
of the angular velocity needs to be revisited.
10.2.3 Angular Velocity of the Spherical Frame
To find a formula for computing the velocity and acceleration directly in the spherical
frame, we follow the same procedure as for polar coordinates in Chapter 3. We start
by taking the derivative of the position of P expressed in the spherical frame:
Id
rP/O = ṙer + r (er ).
dt
dt
Id
(10.23)
I
What is the unit-vector derivative dtd (er )? This question is more difficult to answer
than before because Bs may not simply be rotating about a fixed axis along one of its
unit-vector directions. To find an expression, we turn again to the intermediate frame
A. Because frame A is changing orientation with respect to Bs by rotating only about
the a2 = eθ axis (a simple rotation), we can use our previous results and the unit-vector
form of the transport equation to write the derivative of er with respect to A:
Ad
dt
(er ) = AωBs × er ,
(10.24)
where AωBs = φ̇a2 = φ̇eθ .
Likewise, the transport equation can be used to relate the derivatives between A
and I because the relative orientation between A and I is also only a simple rotation,
this time about ez = a3:
Id
dt
(er ) =
Ad
dt
(er ) + IωA × er ,
(10.25)
where IωA = θ̇ez = θ̇a3. Combining Eqs. (10.24) and (10.25) yields
Id
dt
(er ) = IωA × er + AωBs × er =
ω + AωBs × er .
I A
This result looks just like the unit-vector derivative in Chapter 3. Recall that we
first introduced the angular velocity in Chapter 3 as an operator that computes unitvector derivatives. The same is true here, only now it is the sum of two simple angular
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CHAPTER TEN
velocities. In fact, we have partially verified the addition property of angular velocities
in three dimensions! Introducing the definition
I Bs I A
= ω + AωBs
ω
(10.26)
allows us to write the vector derivative of er as
Id
dt
(er ) = IωBs × er .
(10.27)
A similar development leads to the derivatives of eθ and eφ .
Eq. (10.27) is the same as the planar equation in Eq. (3.32). The only difference is
in the definition of the angular velocity IωBs . In Chapter 3, it was easy to relate the
angular velocity to the geometry of motion, as it corresponded to a simple rotation
about an axis perpendicular to the plane. Here the angular velocity results from
the sum of two simple angular velocities associated with the rotation relative to an
intermediate frame. Nevertheless, it still acts as the appropriate operator to provide
the unit-vector derivative. What we have not (yet) shown is the physical and geometric
interpretations of the three-dimensional angular velocity.
When dealing only with planar rotations, it was clear that the angular velocity
corresponded to the rate of rotation about a particular axis. That is no longer obvious
here. What, then, does the angular velocity defined in Eq. (10.26) represent? You
might have guessed by now that it represents the instantaneous rate of rotation about
an axis aligned with IωBs (we call this the instantaneous axis of rotation). That is, in
fact, true. However, we postpone validating this claim until we treat general frame
rotations. For now, it is sufficient to remember that the angular-velocity addition
property applied to the intermediate frame gives the operator needed to perform unitvector derivatives of the spherical frame.
To perform the cross product in Eq. (10.27), it is convenient to express the angular velocity entirely as components in the Bs frame. From the two simple angular
velocities and Eq. (10.26), we have
I Bs
ω
= φ̇eθ + θ̇a3,
which, using Eq. (10.12), becomes
I Bs
ω
= −θ̇ sin φeφ + φ̇eθ + θ̇ cos φer .
(10.28)
This is the spherical-frame angular velocity. We can substitute Eq. (10.28) into
Eq. (10.27) to get the unit-vector derivatives in terms of the spherical coordinate
angular rates of change (remember the order of terms matters when taking the cross
product). For each of the unit vectors in Bs , we have
Id
dt
Id
dt
Id
dt
(er ) = φ̇eφ + θ̇ sin φeθ
(eθ ) = −θ̇ sin φer − θ̇ cos φeφ
(eφ ) = −φ̇er + θ̇ cos φeθ .
PARTICLE KINEMATICS AND KINETICS IN THREE DIMENSIONS
I
421
P
ez
rP/O
EL
O
ey
ex
AZ
Figure 10.7 Radar tracking antenna. Image courtesy of Shutterstock.
Substituting the er derivative back into Eq. (10.23) gives the velocity
I
vP/O = r φ̇eφ + r θ̇ sin φeθ + ṙer ,
which matches Eq. (10.14). We can take the derivative again and use the other two
angular-velocity cross products to find the acceleration expressed as components in
spherical frame Bs . The result, of course, is the same as in Eq. (10.15).
Example 10.4 Target Tracking and Intercept
Figure 10.7 shows a radar dish antenna. This device is used to track such targets as
an airplane or a missile by sending out radar pulses and timing their return. To point
at the target, it turns toward the maximum strength of the return signal. To find the
range to the target, it computes the round-trip time of each radar pulse. The Doppler
shift of the return pulses provides the range rate.
The dish antenna rotates about two axes to stay pointed at the target: the vertical
axis, with azimuth angle AZ, and the horizontal axis, with elevation angle EL. These
angles are related to the spherical coordinate angles, with EL corresponding to the colatitude (i.e., EL = π/2 − φ). We can imagine a spherical frame fixed to the dish; the
mechanisms that rotate and tilt the dish are physical manifestations of the intermediate
frame. Thus, if encoders on the motors provide the angles, the exact position of the
target relative to the antenna can be found using the spherical-coordinate formula:
rP/O = r cos(AZ) cos(EL)ex + r sin(AZ) cos(EL)ey + r sin(EL)ez .
Likewise, if the range rate and angular rates are provided, then the target velocity can
be found from Eq. (10.14).
Alternatively, we may know the target position and velocity (and acceleration)
from on-board instruments, such as a Global Positioning System (GPS) receiver, and
may want to use that information to command the antenna. Suppose the position
and inertial velocity in Cartesian coordinates (x, y, z)I have been provided. We can
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then find the modified spherical coordinates (AZ, EL, r)I by inverting the spherical
coordinate equations to obtain
r = x 2 + y 2 + z2
z
EL = arcsin
r
y
x
AZ = arcsin
.
= arccos
x2 + y2
x2 + y2
The rates of change of the spherical coordinates in this case are most easily found
by inverting the velocity equation in Eq. (10.14), using IvP/O = ẋex + ẏey + żez and
Eq. (10.13) with θ = AZ and φ = π/2 − EL:
ṙ = ẋ cos(AZ) cos(EL) + ẏ sin(AZ) cos(EL) + ż sin(EL)
˙ = (−ẋ sin(AZ) + ẏ cos(AZ))/(r cos(EL))
AZ
˙ = (−ẋ cos(AZ) sin(EL) − ẏ sin(AZ) sin(EL) + ż cos(EL))/r.
EL
The angular velocity of the frame fixed to the dish is given by Eq. (10.28).
10.3 Linear Momentum, Angular Momentum, and Energy
Now that we know how to write the velocity and acceleration of a particle in three
dimensions, and in cylindrical and spherical coordinates, we need to return to the
material in Chapters 4–7 and generalize it, too, to three dimensions. In those chapters
we always used a vector treatment, specializing to two dimensions only when necessary. As a consequence, our results can be used almost unchanged. For completeness,
however, we review the important ideas.
10.3.1 Linear Momentum and Angular Momentum
Chapter 4 introduced the concept of linear impulse and the resulting momentum
balance in Eq. (4.2), which we repeat here:
mP IvP/O (t2) = mP IvP/O (t1) +
t2
FP dt.
t1
That treatment was completely general and thus translates unchanged to three dimensions. In practice, of course, when writing the velocities in whatever coordinate
system is convenient for the problem at hand, we should be careful to use the new
three-dimensional results.
Likewise for angular momentum and angular impulse. The definition of angular
momentum in Definition 4.2 is unchanged in three dimensions:
rP/O × IpP/O = rP/O × mP IvP/O .
hP/O =
I
PARTICLE KINEMATICS AND KINETICS IN THREE DIMENSIONS
423
The resulting formulation of Newton’s second law is also unchanged, as the definition
of the moment was also a general vector equation:
Id
dt
hP/O = MP/O ,
I
rP/O × FP . The impulse form of the angular momentum equation in
where MP/O =
Eq. (4.17) also follows.
Example 10.5 Solving the Spherical Pendulum
Using Angular Momentum
We now re-solve the spherical pendulum using angular momentum of a particle. This
reduces the number of equations and avoids having to eliminate the reaction force in
the rod, which produces no moment about the attachment point. Using the kinematics
from Example 10.3, the angular momentum of the pendulum bob about the attachment
point O becomes
hP/O = rP/O × mP IvP/O = ler × mP (l φ̇eφ + l θ̇ sin φeθ )
I
= mP l 2φ̇eθ − mP l 2θ̇ sin φeφ .
(10.29)
We now take the inertial derivative of the angular momentum,
Id
dt
Id
Id
hP/O = mP l 2 φ̈eθ + φ̇ eθ − θ̈ sin φeφ − θ̇ φ̇ cos φeφ − θ̇ sin φ eφ ,
dt
dt
I
which, after substituting for the unit-vector derivatives and simplifying, gives
Id
dt
hP/O = mP l 2 φ̈ − θ̇ 2 sin φ cos φ eθ − mP l 2 θ̈ sin φ + 2θ̇ φ̇ cos φ eφ .
I
This equation equals the moment acting on the pendulum bob relative to origin O.
The force that produces a moment about O is gravity, which implies (see Eq. (10.13))
MP/O = rP/O × FP/O = ler × −mP gez
= −mP gler × (− sin φeφ + cos φer )
= mP gl sin φeθ .
Equating the moment on the pendulum bob with the rate of change of angular
momentum yields two scalar equations:
0 = 2mP l 2θ̇ φ̇ cos φ + mP l 2θ̈ sin φ
mP gl sin φ = mP l 2φ̈ − mP l 2θ̇ 2 sin φ cos φ,
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which, after simplifying, result in the same equations of motion as Eqs. (10.21) and
(10.22):
g
sin φ = 0
l
(10.30)
θ̈ + 2θ̇ φ̇ cot φ = 0.
(10.31)
φ̈ − θ̇ 2 sin φ cos φ −
Now we explore the equations of motion a bit more. At first glance they seem less
useful than you might expect because Eq. (10.31) goes to infinity when φ is 0 or π.
This is because, at those angles (the pendulum straight up or straight down), the other
spherical angle, θ, is undefined. However, if you examine Eq. (10.31) closely you will
see it is integrable. Multiplying by sin2 φ gives
θ̈ sin2 φ + 2θ̇ φ̇ sin φ cos φ = 0.
This equation is equal to
d
(mP l 2θ̇ sin2 φ) = 0,
dt
(10.32)
where we have multiplied by mP l 2. Using the transformation table for spherical coordinates in Eq. (10.13) to rewrite the angular momentum in Eq. (10.29) as components
in the inertial frame, we see that the quantity in Eq. (10.32) is the ez component of
the angular momentum of the pendulum bob relative to O. Eq. (10.32) states that the
z-component of the angular momentum is a constant of the motion. This is because
gravity cannot produce a moment about the z-axis.
We can use the fact that hz = IhP/O . ez = mP l 2θ̇ sin2 φ is conserved to rewrite
Eq. (10.30) as
φ̈ −
h2z cos φ
m2p l 4 sin3 φ
−
g
sin φ = 0.
l
Using conservation of one component of the angular momentum, we have again been
able to reduce this two-degree-of-freedom problem to a single differential equation of
motion. A reduction of this sort is often possible in problems that exhibit symmetry—
in this case, the spherical pendulum is symmetric about the ez axis.
10.3.2 Energy of a Particle in Three Dimensions
We now generalize our treatment of work and energy of a particle to three dimensions.
Definition 5.1 is a vector equation, so it still applies in three dimensions; we just
need to be careful about our choice of coordinates. The definition of power remains
unchanged and the work–kinetic-energy formula in Eq. (5.6) remains
TP/O (t2) = TP/O (t1) + WP(tot)(rP/O ; γP ),
where the kinetic energy is
1
1
TP/O = mP IvP/O . IvP/O = mP IvP/O 2.
2
2
PARTICLE KINEMATICS AND KINETICS IN THREE DIMENSIONS
425
The vector definition of a conservative force still holds, as does the treatment
that led us to introduce the potential energy. Thus the final work energy formula in
Eq. (5.16) remains
EP/O (t2) = EP/O (t1) + WP(nc)(rP/O ; γP ),
where the total energy is given by EP/O (t) = TP/O (t) + UP/O (t), and the potential
(F )
energy, for conservative forces, satisfies FP(c) = −∇UP/OP (rP/O ). The important thing
to remember here is that we are now considering the potential energy to be a function
of three spatial coordinates and the gradient operator must be treated accordingly.
(Appendix A shows how to write the gradient of a scalar function in terms of the
various three-dimensional coordinate systems.)
Example 10.6 The Energy of a Spherical Pendulum
We now look at the energy of a spherical pendulum. Since this example contains a
single particle, albeit with two degrees of freedom, the kinetic energy is calculated
using the velocity from Example 10.3:
1
1
TP/O = mP IvP/O 2 = mP l 2(φ̇ 2 + θ̇ 2 sin2 φ).
2
2
The potential energy is a function of the height of the pendulum:
UP/O = mP grP/O . ez ,
which, using spherical coordinates and the transformation table (Eq. (10.13)), gives
UP/O = mP gl cos φ.
The total energy is the sum of the kinetic and potential energies:
1
EP/O = mP l 2(φ̇ 2 + θ̇ 2 sin2 φ) + mP gl cos φ.
2
Because the only force doing work here is gravity, which is conservative, the total
energy of the pendulum is conserved.
We can further simplify the expression for the energy. Recall that Example 10.5
showed that the component of the angular momentum of the pendulum bob about the
ez axis is a constant of the motion:
hz = mP l 2θ̇ sin2 φ = constant.
This result can be used to solve for θ̇, which we then substitute into the energy
expression. After a bit of algebra, we find
h2z
1
EP/O = mP l 2φ̇ 2 +
+ mP gl cos φ,
2
2mP l 2 sin2 φ
which is a function of φ and φ̇ only.
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CHAPTER TEN
10.4 Relative Motion in Three Dimensions
This section generalizes to three dimensions our treatment of relative motion in Chapter 8. Unlike in the previous section, where the treatment of particles in two dimensions transferred directly, here the situation becomes quite a bit more complicated.
Recall what we set out to do in Chapter 8. Our goal was to consider a body frame B
rotating and translating with respect to an inertial frame I . Point P was allowed to
move arbitrarily in B, and we described its inertial velocity and inertial acceleration
in terms of its motion in B. The only restriction was that B changed its orientation
in I by means of a simple rotation about the b3 = e3 axis; the motion of P was confined to a plane. The result was the transport equation and the acceleration equation
in Eq. (8.21), which we have used frequently since.
The restriction to a simple rotation between B and I was important. It allowed
us to derive, using the transformation table, the formula for the time derivative of a
body-frame unit vector in absolute space. Now we need to generalize this derivation
to an arbitrary rotation between B and I . The basic approach is the same as in
Section 10.2.3—we will use intermediate frames. In fact, we just need one more
rotation and one more intermediate frame.
10.4.1 Orientation Angles
If you review Chapter 3, particularly Section 3.4, you will see that before introducing
the polar and path frames and using them to develop the kinematics of a particle
moving in the plane, we first introduced the idea of relative orientation and the
transformation table. We have used that idea frequently since. In particular, it was
essential for the development of the transport equation (Definition 8.1). There we
considered only three-degree-of-freedom problems, two for translation and one for
rotation, so a single angle was sufficient to describe the orientation of the body frame
in absolute space.
Now things get a bit more complicated, but we follow the same general approach.
In three dimensions, a rigid body has, in general, six degrees of freedom—three
for translation and three for rotation. It takes three scalar coordinates to completely
describe the orientation of a rigid body (or body frame) in absolute space. Our first
task, then, before turning to the kinematics of a particle or rigid body, is to find a way
of describing the general three-dimensional orientation of a body frame (or a rigid
body) relative to another frame using three scalar coordinates.
There are, as might be expected, many ways to describe the orientation of a frame,
say B, with respect to a second frame, say I (i.e., there are many sets of three scalar
coordinates), just as there are many scalar coordinates that can be used to describe
the position of a point in a reference frame. Our discussion of the spherical frame
leads us to think of a particular approach—using simple rotations about the axes
of intermediate frames. For the spherical frame, we had two angles, θ and φ, that
oriented the frame. For a general body frame B, three angles are needed to describe
the orientation of B in I . The most common set (and notation) are angles ψ, θ, and
φ, which involve two intermediate frames. In similar fashion to the spherical frame,
PARTICLE KINEMATICS AND KINETICS IN THREE DIMENSIONS
I
427
A
a3
e3, a3
C
c3
A
a2
a2,c2
e2
ψ
a1
c1
e1
(a)
θ
a1
(b)
C
c3, b3
b2
c2
c1
ϕ
b1
(c)
Figure 10.8 A 3-2-3 ordered rotation via the Euler angles (ψ, θ, φ)IB of frame B in frame
I using intermediate frames A and C. (a) ψ rotation about e3. (b) θ rotation about a2 . (c) φ
rotation about c3.
we first rotate about the e3 axis of frame I by an angle ψ to get to the intermediate
frame A. We then rotate by θ about the a2 axis of frame A to reach the intermediate
frame C. (This frame is the same as the spherical frame!) Finally, we rotate about the
c3 axis by φ to get to the B frame. Note that all are right-handed rotations.
The set of angles (ψ, θ, φ)IB are known as orientation angles or Euler angles.(Note
the slight change in notation when the scalar coordinates describe the orientation
of B in I rather than the location of a particle relative to an origin; we need to
indicate both frames.) The particular ordered set of rotations described above and
shown in Figure 10.8 is known as a 3-2-3 rotation, as it involves an ordered sequence
of rotations about the 3-axis, the 2-axis, and the 3-axis of a series of intermediate
reference frames. It is possible to describe any orientation of reference frame B in
I by means of these three angles. Thus they constitute a set of scalar coordinates
for describing the orientation of a reference frame. In fact, you have already seen an
example of using Euler angles in Example 10.4, where we used azimuth (AZ) and
elevation (EL) to describe the orientation of a radar tracking dish.
As you might have guessed, there are also many other possible sets of Euler angles
(e.g., a 3-1-3 rotation or a 1-2-3 rotation). In all, there are 12 possible sets of three
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Euler-angle rotations that can be used to describe the orientation of one frame with
respect to another. Which set is best depends on the specific mechanical system being
analyzed. We use a few different sets in the examples that follow.
We begin by examining how to transform a vector from one set of unit vectors to
another set in terms of the Euler angles, just as in the planar case. That is, how do we
write the unit vectors of frame B in terms of those in I ? We use the same approach
as for the spherical frame and write the transformation tables between each pair of
intermediate frames related by a simple rotation. Thus the transformation between
the inertial frame I and the first intermediate frame A is
a1
a2
a3
e1
e2
cos ψ
− sin ψ
0
sin ψ
cos ψ
0
e3
0
0
1 .
The second transformation between A and C is
c1
c2
c3
a1
a2
a3
cos θ
0
sin θ
0
1
0
− sin θ
0
cos θ ,
and the transformation between C and B is
b1
b2
b3
c1
c2
cos φ
− sin φ
0
sin φ
cos φ
0
c3
0
0
1 .
These transformation tables allow us to write down the unit vectors of B as a sum
of components in each frame, and eventually in frame I . Thus b1 is
b1 = cos φc1 + sin φc2
= cos φ cos θa1 + sin φa2 − cos φ sin θa3
= (cos φ cos θ cos ψ − sin φ sin ψ)e1 + (cos φ cos θ sin ψ + sin φ cos ψ)e2
− cos φ sin θe3.
Likewise, solving for b2 and b3 as components in I yields
b2 = −(cos ψ cos θ sin φ + cos φ sin ψ)e1 − (sin ψ cos θ sin φ − cos φ cos ψ)e2
+ sin θ sin φe3
b3 = cos ψ sin θe1 + sin ψ sin θe2 + cos θe3.
As for the spherical frame, we can consolidate the transformations into a table
relating frame I and frame B:
PARTICLE KINEMATICS AND KINETICS IN THREE DIMENSIONS
e1
b1
b2
b3
e2
429
e3
cos φ cos θ cos ψ − sin φ sin ψ
cos φ cos θ sin ψ + sin φ cos ψ − cos φ sin θ
− cos ψ cos θ sin φ − cos φ sin ψ − sin ψ cos θ sin φ + cos φ cos ψ sin θ sin φ
cos ψ sin θ
sin ψ sin θ
cos θ
.
(10.33)
Eq. (10.33) contains a three-dimensional transformation that relates the unit vectors of
the inertial frame to those of an arbitrary body frame B whose orientation is described
by a set of 3-2-3 Euler angles (ψ, θ, φ)IB . As in the planar case, the row i, column j
element is given by bi . ej .
It is also convenient to describe these unit-vector transformations by means of
matrix operations. We thus conclude this section with a rederivation using matrix
notation. We have not used this notation very much, but it becomes very helpful here
and in the next chapter.
Recall that Chapter 3 introduced a matrix notation for writing the components of
a vector in a frame, in which the magnitudes of the components are the elements of a
column matrix and the unit vectors are implied by the subscript. Thus a vector r written
as components in frame B as r = xb1 + yb2 + zb3 is written in matrix notation as
⎡
⎤
x
[r]B = ⎣ y ⎦ .
z B
(10.34)
Suppose we use the transformation table to express r in frame C. Its components
are then given by r = (x cos φ − y sin φ)c1 + (x sin φ + y cos φ)c2 + zc3, which in
matrix form is
⎤
⎡
⎤
x cos φ − y sin φ
r1
[r]C = ⎣ r2 ⎦ = ⎣ x sin φ + y cos φ ⎦ .
r3 C
z
C
⎡
(10.35)
Comparing Eq. (10.35) with Eq. (10.34) shows that [r]B and [r]C are related by a
simple matrix multiplication:
⎡
⎤
⎡
r1
cos φ
⎣ r2 ⎦ = ⎣ sin φ
r3 C
0
− sin φ
cos φ
0
⎤⎡ ⎤
x
0
0⎦⎣y⎦ .
z B
1
(10.36)
=CC B
The matrix CC B defined in Eq. (10.36) is the transformation matrix in three dimensions, also called the direction-cosine matrix. It acts to transform the components of
a vector from one frame to another using matrix multiplication.
Note that we haven’t really done anything new here. The elements of CC B are
just the elements of the corresponding transformation table. The value of writing
the transformation table as a matrix is that it converts many vector operations to
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CHAPTER TEN
linear algebra. For instance, if we write the other two intermediate transformations as
matrices,
⎡
⎤
cos θ
0 sin θ
A C
C =⎣ 0
1
0 ⎦
− sin θ 0 cos θ
and
⎡
cos ψ
I A
C = ⎣ sin ψ
0
− sin ψ
cos ψ
0
⎤
0
0⎦,
1
then the net transformation between frames I and B is given by the matrix multiplication IC B = IC AAC CCC B , which yields
⎡
cos φ cos θ cos ψ − sin φ sin ψ
I B
⎣
C = cos φ cos θ sin ψ + sin φ cos ψ
− cos φ sin θ
− cos ψ cos θ sin φ − cos φ sin ψ
− sin ψ cos θ sin φ + cos φ cos ψ
sin θ sin φ
⎤
cos ψ sin θ
sin ψ sin θ ⎦.
cos θ
(10.37)
Turning successive rotations of frames into a series of matrix multiplications is a
general property of the transformation matrix. The notation we have chosen, which
explicitly notes the starting and ending frames, helps remind you of the proper order of
multiplications. It is also still true that the elements of IC B are given by Cij = bi . ej ,
i,j = 1, 2, 3. This explains why it is called the direction-cosine matrix: each element
is the cosine of the angle between the corresponding pair of unit vectors.
Compare the expression for IC B to our old notation using the transformation
table in Eq. (10.33). This should convince you that, even for this more complicated
matrix including all three rotations, it is still true that IC B = (BC I )T . In other words,
we have
[r]I = IC B [r]B
(10.38)
[r]B = BC I [r]I = (IC B )T [r]I
(10.39)
and
for any transformation matrix IC B . Of course, because these equations are simple
matrix equations, we should have been able to get from Eq. (10.38) to Eq. (10.39)
simply by multiplying on the left by the inverse of IC B . This is true, and it implies that
(IC B )−1 = (IC B )T or, equivalently, that IC B (IC B )T = I , where I is the 3 × 3 identity
matrix. Matrices that satisfy this property are called orthogonal. Orthogonality is a
general property of all transformation matrices.
Finally, though we won’t prove it, a transformation matrix also has the property
|IC B | = 1 and |IC B − I | = 0, where the notation | . | represents the matrix determinant. These two properties imply that one of the eigenvalues of the transformation
matrix is always unity. Thus there always exists a vector (an eigenvector of IC B )
such that
PARTICLE KINEMATICS AND KINETICS IN THREE DIMENSIONS
431
⎡
⎤
⎡ ⎤
c1
c1
⎣ c2 ⎦ = IC B ⎣ c2 ⎦ .
c3 I
c3 B
We use this property later when discussing angular velocity in three dimensions.
Example 10.7 Gimbals
Example 10.4 described the orientation of the body frame attached to a radar dish
by what we now know is a 3-2-3 orientation set (though the third angle in the set
was not used). Even though we did not yet have a transformation table, we were
nonetheless able to describe the position of a target fixed in the body frame in terms of
components in the inertial frame using two Euler angles (AZ and EL) and its distance
from the origin. The mechanism pointing the dish is a physical representation of the
intermediate frame used to define the spherical body frame.
This example describes gimbals, another common mechanical system used to
manifest a set of sequential Euler angles. You may be familiar with gimbals from
a visit to an amusement park. A very popular ride, shown in Figure 10.9, involves
strapping the patron into the center gimbal and allowing her to rotate freely to any
arbitrary orientation. Gimbals date back to ancient Greece and have been used for
millennia to keep objects oriented with the vertical. In modern times they have found
their most common use in inertial navigation systems, where they support a gimbaled
gyroscope. Chapter 11 covers much more about gyroscopes; for now, you only need
to know that, in the absence of torques, the orientation of a spinning gyroscope stays
fixed in absolute space.
Figure 10.9 Amusement park ride using three gimbals.
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CHAPTER TEN
Intermediate
axis
ψ
I
e1
ϕ
B
b1
e2
θ
e3
b3
(a)
b2
(b)
Figure 10.10 (a) Gimbaled gyroscope. (Made by Educational Control Products-ECP.)
(b) Gimbal schematic.
Figure 10.10a shows a picture of a gimbaled gyroscope. By attaching the outer
gimbal to a vehicle and making the joints free of friction, the orientation of the
gyroscope stays fixed in absolute space while the vehicle (e.g., airplane, submarine,
or spacecraft) moves around it. Thus using encoders to measure the angles of each
gimbal directly provides the Euler angles that describe the orientation of a vehicle
body frame relative to the gyro body frame (which is an inertial frame).
A schematic drawing of a typical set of gimbals is shown in Figure 10.10b. The
orientation of the central disk relative to absolute space is represented by a 3-1-3
rotation set. We again use the Euler angles (ψ, θ, φ)IB to represent the orientation of B
in I , only now ψ represents a rotation about the inertial 3-axis, θ represents a rotation
about the intermediate 1-axis, and φ represents a rotation about the final 3-axis.
Vectors in the body frame (e.g., the location of the rider’s head) can be transformed
to components in the inertial frame by means of a transformation table (or directioncosine matrix) found in a similar way to the 3-2-3 one in Eq. (10.33). The resulting
3-1-3 transformation table is
e1
e2
e3
b1
− sin φ cos θ sin ψ + cos φ cos ψ sin φ cos θ cos ψ + cos φ sin ψ sin φ sin θ
b2
− sin ψ cos θ cos φ − sin φ cos ψ cos ψ cos θ cos φ − sin φ sin ψ sin θ cos φ
b3
sin ψ sin θ
− cos ψ sin θ
cos θ
.
Example 10.8 The Orientation of an Airplane
The airplane is a classic example of a three-dimensional rigid body undergoing
translation and rotation as it flies from point to point and changes orientation (perhaps
pitching to increase angle of attack or rolling and yawing to enter a turn). It is thus
an excellent example of rigid-body motion, as there are no constraints; it has the full
six degrees of freedom. As shown in the next chapter, the separation principle still
holds in three dimensions, and we can examine the translational motion of the airplane
PARTICLE KINEMATICS AND KINETICS IN THREE DIMENSIONS
b1
433
b2
B
b3
ex
O
ey
ez
I
Figure 10.11 The orientation of an airplane body frame relative to the inertial frame is
described by a 3-2-1 Euler angle set (ψ, θ, φ)IB .
as if it were a point at its center of mass and separately solve for its orientation. We
postpone the treatment of airplane dynamics until later, but here we show the common
description of the orientation of an airplane in the inertial frame.
Figure 10.11 shows a diagram of a typical aircraft with an attached body frame
located at the center of mass of the plane and a fixed inertial frame. We assume the
inertial frame is on the surface of the earth and ignore the curvature and rotation of the
earth. (Yes, the inertial z-axis points down so that an airplane at altitude has a negative
ez component.) By convention the b1 axis of the body frame is along the centerline of
the fuselage and points through the nose of the aircraft. The b3 axis points downward,
so that the plane maintains symmetry on either side, and the b2 axis points to the
right (looking forward) through the wing to complete the right-handed set.
Also by convention, the orientation of the airplane body frame relative to the fixed
inertial frame is described by the ordered 3-2-1 set of Euler angles (ψ, θ, φ)IB , where
the first rotation, ψ, is a yaw rotation about the inertial 3-axis, the second rotation,
θ, is a pitch rotation about the intermediate body 2-axis, and the third rotation, φ,
is a roll rotation about the body 1-axis. We can again find the transformation table
(or direction-cosine matrix) to convert a vector in the body frame into components in
absolute space by means of the same procedure using intermediate frames. The result
is the following transformation table:
ex
ey
ez
b1
cos ψ cos θ
sin ψ cos θ
− sin θ
b2
cos ψ sin θ sin φ − cos φ sin ψ
sin ψ sin θ sin φ + cos φ cos ψ
cos θ sin φ
b3
cos φ sin θ cos ψ + sin φ sin ψ
sin ψ sin θ cos φ − sin φ cos ψ
cos θ cos φ .
10.4.2 The Angular Velocity and the Transport Equation
in Three Dimensions
We now have the tools to generalize the concept of angular velocity and re-derive the
transport equation in three dimensions, thus completing our description of particle
kinematics and kinetics. We repeat the derivation allowing for an arbitrary change of
orientation of the body frame B in I . The procedure is very similar to the treatment in
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CHAPTER TEN
Section 10.2.3, where we calculated the rate of change of the unit vectors associated
with the spherical frame. The consequence, as before, is the introduction of the angular
velocity as an operator used to find the vector derivative.
We begin by taking the inertial derivative of the position rP/O = xb1 + yb2 + zb3,
written as components in body frame B:
Id
Id
Id
rP/O = ẋb1 + ẏb2 + żb3 +x (b1) + y (b2) + z (b3). (10.40)
dt
dt
dt
dt
Id B
= dtd (rP/O )
Once again, we need an expression for the unit-vector derivatives. As for the
spherical frame, the trick is to use intermediate frames so that each incremental
rotation is planar. Let us consider the 3-2-3 rotation with Euler angles (ψ, θ, φ)IB
described in the previous section. We can then use the transport equation already
derived for a simple angular velocity (planar rotation) to relate the derivative of the
unit vector bi in I to its derivative in the intermediate frame A:
Id
dt
(bi ) =
Ad
(bi ) + IωA × bi ,
dt
(10.41)
where IωA = ψ̇a3, and i = 1, 2, 3. We can do the same for frames A and C:
Ad
dt
(bi ) =
Cd
(bi ) + AωC × bi ,
dt
(10.42)
where AωC = θ̇c2. Finally, we use the transport equation one last time between frames
C and B to obtain
Cd
dt
(bi ) =
Bd
(bi ) +CωB × bi ,
dt
(10.43)
=0
where CωB = φ̇b3. The derivative in B is zero because bi is a unit vector fixed in
B. Combining Eqs. (10.41)–(10.43) gives the following expression for the inertial
derivative of bi :
Id
dt
(bi ) = IωA × bi + AωC × bi + CωB × bi
=
ω + AωC + CωB ×bi .
I A
(10.44)
=IωB
We define the angular velocity IωB as the sum of the simple angular velocities
associated with the intermediate frames, just as for the spherical frame. As in the
planar case, the vector quantity we call the angular velocity is an operator that provides
unit-vector derivatives. We have also shown that the angular velocity can be found by
summing the simple angular velocities of successive rotations. This is a generalization
of the addition property established in Section 8.2.2, but here the successive rotations
need not be co-axial. We still have not discussed what the angular velocity represents
PARTICLE KINEMATICS AND KINETICS IN THREE DIMENSIONS
435
physically; because the three intermediate rotations are not co-axial, the physical
interpretation of IωB may not be obvious. Nevertheless, as noted in the discussion of
the spherical frame, it is a fact that this angular velocity represents an instantaneous
rate of rotation about an axis along the vector direction.
We can now substitute the unit-vector derivative contained in Eq. (10.44) back into
Eq. (10.40) to find
Bd rP/O =
rP/O + x IωB × b1 + y IωB × b2 + zIωB × b3,
dt
dt
Id or
Bd rP/O =
rP/O + IωB × rP/O ,
dt
dt
(10.45)
ω = IωA + AωC + CωB = ψ̇a3 + θ̇c2 + φ̇b3.
(10.46)
Id where
I B
Thus the transport equation is valid in three dimensions, as long as we understand
how to find IωB !
Example 10.9 Airplane Kinematics
This example continues our study of the airplane. At first glance, describing the
translational motion of the airplane in the inertial frame (see Figure 10.12) and finding
the equations of motion using Newton’s second law are easy. We treat the airplane
as a point mass located at its center of mass G and write its position and velocity as
components in I in terms of Cartesian coordinates:
rG/O = xe1 + ye2 + ze3
I
vG/O = ẋe1 + ẏe2 + że3.
b1
b2
B
I
vG/O
rG/O
I
b3
ex
O
ey
G
ez
Figure 10.12 Position and velocity of an airplane with body frame B in inertial frame I .
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The equations of motion are three second-order differential equations in terms of
the forces on the aircraft. However, the translational motion of an airplane is almost
never described this way. We do use Cartesian coordinates to describe its position, but
typically the velocity is expressed in terms of components in the body frame rather
than the inertial frame. This is because most of the forces and moments on the plane
are due to aerodynamics and are functions of the orientation of the velocity in the
body frame of the aircraft. For instance, the angle that the velocity makes with the
body b1 axis (through the nose) is called the angle of attack and the lift on the plane
is directly proportional to it. It is thus much more convenient to calculate the lift by
working with the velocity expressed in terms of components in B.
There are two approaches for doing this. The first is by means of the transport
equation. That approach involves first expressing the position of the plane’s center of
mass as components in the body frame, which is most easily done in matrix notation.
We have
⎡ ⎤
x
[rG/O ]B = B C I ⎣ y ⎦ ,
z I
where the direction-cosine matrix B C I is derived from the transformation table for
the 3-2-1 Euler angles in Example 10.8. We then use the transport equation (in matrix
form) to find the components of the inertial velocity in the body frame,
⎛
⎡ ⎤ ⎞
x
Bd
⎜B I ⎣ ⎦ ⎟
I B
I
[ vG/O ]B =
y
⎠ + ω × rG/O ,
⎝ C
B
dt
z I
where an airplane’s angular-velocity components in body axes are commonly given
by the letters p, q, and r:
I B
ω = pb1 + qb2 + rb3.
Although this approach certainly works to provide the velocity in terms of components in the body frame and has the added advantage of expressing the velocity
components as functions of the rate of change of Cartesian coordinates, the process
is tedious, and the resulting expression is extremely complex, particularly if we try to
replace the rates of change of the Euler angles with the angular-velocity components.
Instead, standard practice is to simply introduce new variables for the component
magnitudes of the velocity along body axes:
I
vG/O = ub1 + vb2 + wb3.
When we find the dynamical equations of motion later using Newton’s second law, we
do so in terms of these three scalars (u, v, and w). This approach greatly reduces the
amount of algebra needed to derive the airplane equations of motion. (Remember, we
need the velocity expressed in the body frame to determine the forces and moments.)
The disadvantage is that we are no longer describing the translational state of the
airplane center of mass by coordinates and their rates but rather by the coordinates
and the component magnitudes of the velocity in the body frame, which are functions
PARTICLE KINEMATICS AND KINETICS IN THREE DIMENSIONS
437
of the coordinate rates. In fact, the kinematic equations relating the rate of change of
the coordinates to the new velocity variables are given by the transformation
⎡
⎤
⎡ ⎤
u
ẋ
[IvG/O ]I = ⎣ ẏ ⎦ = IC B ⎣ v ⎦ ,
w B
ż I
(10.47)
where
⎡
cos θ cos ψ
⎢
C = ⎣ sin ψ cos θ
I B
− sin θ
cos ψ sin θ sin φ − cos φ sin ψ
sin ψ sin θ sin φ + cos φ cos ψ
cos ψ sin θ cos φ + sin φ sin ψ
⎤
⎥
sin ψ sin θ cos φ − cos ψ sin φ ⎦ .
cos θ sin φ
cos θ cos φ
Thus, given trajectories u(t), v(t), and w(t) for the velocity and Euler angles ψ(t),
θ (t), and φ(t) (found from solving the equations of motion, done later in this chapter
and the next), Eq. (10.47) can be integrated to find the inertial position of the plane.
We further discuss the implications of this change of variables later in the chapter.
10.4.3 Angular Velocity and Euler-Angle Rates
Our last step in the kinematics of three-dimensional rotation is to relate the components of the angular velocity of the body frame B in I to the rates of change of the
Euler angles, as these are the scalar coordinates used to describe the orientation of B
in I . Recall that, for the planar case, where the angular velocity is simple, this calculation was straightforward. The angular-velocity magnitude was simply θ̇, the rate
of change of rotation of the body frame. In three dimensions, however, it gets more
complicated.
From Eq. (10.46), we already have
I B
ω = ψ̇a3 + θ̇c2 + φ̇b3
for the 3-2-3 rotation. The problem is that this expression is written in terms of unit
vectors of three different frames. It is most convenient to have an expression for IωB in
terms of components in B when applying the transport equation. This is accomplished
using the 3-2-3 transformation tables in Section 10.4.1, which yield
I B
ω = −ψ̇ sin θc1 + ψ̇ cos θc3 + θ̇c2 + φ̇b3
= (θ̇ sin φ − ψ̇ sin θ cos φ)b1 + (θ̇ cos φ + ψ̇ sin θ sin φ)b2 + (φ̇ + ψ̇ cos θ)b3.
If we write the components of the angular velocity IωB = ω1b1 + ω2b2 + ω3b3 then
the expression for IωB is equivalent to the three scalar equations
ω1 = θ̇ sin φ − ψ̇ sin θ cos φ
ω2 = θ̇ cos φ + ψ̇ sin θ sin φ
ω3 = φ̇ + ψ̇ cos θ.
(10.48)
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These are called the kinematic equations of rotation. They relate the components
of the angular velocity to the rate of change of the scalar coordinates describing
orientation, just as there are equations that relate the components of the velocity
of a particle to the rate of change of the scalar coordinates describing the position
of the particle (e.g., Cartesian, cylindrical, or spherical coordinates). We thus have
a parallel representation of the kinematics of three-dimensional orientation to that
of three-dimensional translation. Our goal in the next chapter is to develop the
dynamics of a rigid body, represented by a body frame B. The resulting equations
of motion are differential equations for the components of the angular velocity that
we can solve simultaneously with the kinematic equations of rotation to find complete
rotational trajectories. Just as with translation, each of the three degrees of freedom
associated with rotation has a position-like coordinate (the Euler angles) and a speedlike coordinate (the angular-velocity components). Alternatively, we can substitute
from Eq. (10.48) into the equations of motion for IωB to find three second-order
differential equations for the Euler angles, as we did for the other coordinates earlier
in the book.
You probably noticed that, if we have integrable equations of motion for the rates
of change of ω1, ω2, and ω3, then we also need the inverse of the kinematic equations
of rotation to solve for the Euler angles (ψ, θ, φ)IB . Fortunately, these equations are
not difficult to find. Simple algebra yields
ψ̇ = (−ω1 cos φ + ω2 sin φ) csc θ
θ̇ = ω1 sin φ + ω2 cos φ
(10.49)
φ̇ = (ω1 cos φ − ω2 sin φ) cot θ + ω3.
You may also notice that these equations have a problem: at θ = 0 or π they are
singular (the right-hand side goes to infinity). This is a well-known problem with
the Euler angle description of orientation and is referred to as gimbal lock. It occurs
because there is an ambiguity between ψ and φ for describing the orientation of B in
I when θ = 0. (This problem came up in Example 10.3 for the spherical pendulum.)
We call it gimbal lock because the singular value of θ corresponds to alignment of two
of the gimbals (see Example 10.7), which creates a measurement ambiguity. Gimbal
lock is one of the reasons that another set of Euler angles might be used for a particular
problem. It is beyond our scope to explore this issue in more detail. The notes to this
chapter suggest a number of texts for further reading.
Example 10.10 Kinematic Equations of Rotation for 3-1-3 and 3-2-1
Euler Angles
Examples 10.7 and 10.8 introduced two alternative sets of Euler angles convenient
for specific geometries: the 3-1-3 angles for a set of gimbals and the 3-2-1 angles
describing the orientation of an airplane. We also defined the transformation table (or
direction-cosine matrix) for each. Just as they have different transformation tables,
they also have different kinematic equations of rotation to relate the rates of change
of the Euler angles to the angular-velocity components in the body frame. Thus,
following the procedure above, the angular velocity for the 3-1-3 set is
I B
ω = ψ̇a3 + θ̇c1 + φ̇b3.
PARTICLE KINEMATICS AND KINETICS IN THREE DIMENSIONS
439
Using the appropriate frame transformations gives the 3-1-3 kinematic equations of
rotation:
ω1 = θ̇ cos φ + ψ̇ sin θ sin φ
ω2 = −θ̇ sin φ + ψ̇ sin θ cos φ
ω3 = φ̇ + ψ̇ cos θ.
Likewise, for the 3-2-1 set we have
I B
ω = ψ̇a3 + θ̇c2 + φ̇b1.
The frame transformations lead us to the 3-2-1 kinematic equations of rotation,
ω1 = p = φ̇ − ψ̇ sin θ
ω2 = q = θ̇ cos φ + ψ̇ cos θ sin φ
ω3 = r = −θ̇ sin φ + ψ̇ cos θ cos φ,
where we have added the airplane notation for the angular-velocity components as a
reminder (see Example 10.9).
Example 10.11 The Barrel Roll and Other Maneuvers
This example examines how to use the inverse relationship between the angularvelocity components (p, q, and r) of the airplane and the Euler angle rates (ψ̇, θ̇,
and φ̇). Inverting the 3-2-1 kinematic equations from the previous example yields
ψ̇ = (q sin φ + r cos φ) csc θ
θ̇ = q cos φ − r sin φ
(10.50)
φ̇ = (q sin φ + r cos φ) tan θ + p.
Imagine the airplane performing a barrel roll about the b3 axis, as shown in
Figure 10.13. During such a maneuver, q = r = 0, and p = 0. Eq. (10.50) tells us
that ψ̇ = θ̇ = 0 and φ̇ = p, perhaps not a surprise. However, if the airplane performs
a loop de loop maneuver, during which p = r = 0 and q = 0, observe that
ψ̇ = q sin φ csc θ
θ̇ = q cos φ
φ̇ = q sin φ tan θ.
That is, the loop de loop maneuver generates nonzero rates of change of all three Euler
angles! In fact, it is only when the airplane loops with exactly zero roll (φ = 0) that
the expected result of ψ̇ = φ̇ = 0 and θ̇ = q holds. Therefore, it is in general a mistake
to assume that simple three-dimensional rotations of a rigid body are associated with
simple angular rates. However, simple rotations are associated with simple angular
velocities, provided that the angular velocity is expressed as components in the body
frame.
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CHAPTER TEN
Figure 10.13 Barrel roll. Image courtesy of Shutterstock.
Example 10.12 Target Tracking, Part 2
This example revisits the target-tracking problem explored in Example 10.4. Here we
study the kinematics using Euler angles and the angular velocity rather than spherical
coordinates. Figure 10.14 shows the dish antenna in the inertial frame, now with a
body frame added and b2 pointing along the symmetry axis of the dish.
In this problem we solve for the commanded rates of change of the Euler angles so
that the dish stays pointed at target P , traveling at constant velocity in the ex direction,
Iv
P/O = vex . For example, P may be a ship with known position and velocity that
needs to maintain communication with a land base via the dish.
The rotation taking the dish from the inertial frame to the arbitrary body-frame
orientation is given by the 3-1-3 Euler angle set with ψ = AZ, θ = EL, and φ = 0.
(No final rotation is needed because of the symmetry of the dish.) The goal is to find
first-order differential equations for the Euler angles that can be solved to determine
the angular-rate commands for the dish motors.
I
B
P
ez
I
b3
vP/O
b2
O
ey
ex
b1
Figure 10.14 Dish antenna with attached body frame B = (O, b1, b2 , b3) tracking point P
traveling at velocity IvP/O . Image courtesy of Shutterstock.
PARTICLE KINEMATICS AND KINETICS IN THREE DIMENSIONS
441
We begin by assuming that the target has position rP/O . If the distance to the target
is d, then
rP/O = db2.
We use the transport equation to rewrite the target inertial velocity,
vP/O = BvP/O + IωB × rP/O = vex ,
I
where BvP/O = ḋb2, the range rate in body axes, and IωB = ω1b1 + ω2b2 + ω3b3.
Writing this out as components in B gives
vex = ḋb2 + ω1db3 − ω3db1.
Using the transformation table derived in Example 10.7 for the 3-1-3 Euler angles
allows us to convert this to the following three scalar equations (by writing ex in body
axes):
v cos ψ = −ω3d
−v sin ψ cos θ = ḋ
v sin ψ sin θ = ω1d,
where we have set φ = 0. These equations are solved for ω1 and ω3, which are then substituted into the kinematic equations for a 3-1-3 Euler angle set from Example 10.10.
Inverting the angular-velocity relationship for the 3-1-3 system gives
ψ̇ = (ω1 sin φ + ω2 cos φ) sec θ
θ̇ = ω1 cos φ − ω2 sin φ
φ̇ = −(ω1 sin φ + ω2 cos φ) cot θ + ω3.
Setting φ̇ = φ = 0 allows us to use the third equation to solve for ω2 = ω3 tan θ. The
three angular velocities are then substituted into the first two equations to give the
following set of differential equations:
v cos ψ
d cos θ
v sin ψ sin θ
θ̇ =
d
ψ̇ = −
ḋ = −v sin ψ cos θ.
Integrating these equations from a known set of initial conditions (e.g., when the
antenna locks onto a target) provides the Euler-angle commands to track the target.
10.4.4 Multiple Frames and the Addition Property
Although it seems like we have already demonstrated the addition property of angular
velocities in three dimensions in Section 10.4.2, that is not quite true, as each angular
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CHAPTER TEN
velocity in the sum was a simple angular velocity associated with an intermediate
frame. Nevertheless, the general addition property of angular velocities in three
dimensions is in fact true. That is, suppose we have three frames undergoing arbitrary
rotations with respect to one another. In other words, suppose that frame A is rotating
in frame I with angular velocity IωA . Suppose also that frame B is rotating in frame
A with angular velocity AωB , and suppose frame C is rotating in frame B with angular
velocity BωC . Then it is a fact that
I C
ω = IωA + AωB + BωC .
(10.51)
This addition property, of course, applies to any number of frames.
Fortunately, we have already derived Eq. (10.51)! Review Section 8.2.2 and Figure 8.9. The derivation there is entirely in terms of vectors and uses only the transport
equation. Because we have now shown that the transport equation is true in three dimensions, the treatment in Section 8.2.2 is completely general and applies in three
dimensions as well. Thus not only is the addition property true for general angular velocities in three dimensions, but the equation for the velocity of point P in I involving
multiple frames of reference is still
I
vP/O = AvO /O + IωA × rO /O + BvO
/O
+ IωB × rO
/O
+ CvP/O + IωC × rP/O .
Note the origin of frames I and A is O, the origin of frame B is O , and frame C has
origin O .
10.4.5 Three-Dimensional Particle Kinetics in a Rotating Frame
If you look back at Section 8.3, where the acceleration of a particle relative to a translating and rotating frame is discussed, you will see that we never used the planar
assumption. Because we have just shown that the transport equation applies in three
dimensions, the entire development in Section 8.3 holds. Nevertheless, for completeness, and to remind you yet again of their importance, we restate the fundamental
equations for velocity and acceleration involving translating and rotating frames. That
is, if frame B is translating and rotating with respect to frame I , as shown in Figure 10.15, then the transport equation can be used to write the velocity of point P in
I in terms of its velocity in B and the velocity of the origin O of frame B:
vP/O = IvO /O + BvP/O + IωB × rP/O .
I
Likewise, using the same derivation as in Section 8.3, the acceleration of P is
I
aP/O = IaO /O + BaP/O + IαB × rP/O + 2IωB × BvP/O
+ IωB × IωB × rP/O .
(10.52)
PARTICLE KINEMATICS AND KINETICS IN THREE DIMENSIONS
443
FP
I
B
e3
P
b3
rP/O′
rP/O
b2
O′
b1
O
e1
e2
Figure 10.15 Particle P subject to force FP moving relative to frame B, which may rotate or
translate in three dimensions relative to frame I .
Just as we did earlier in the book, we can now apply Newton’s second law to
a particle (or collection of particles), given only their velocities and accelerations
in a translating and rotating frame. Our discussion of the centripetal and Coriolis
accelerations in Section 8.3 is unchanged, though we can now treat slightly more
complex examples.
Example 10.13 Tool Bag in Space
This example uses Eq. (10.52) to study the motion of an object relative to a moving
reference frame. Imagine that an astronaut on a space walk outside a spacecraft
accidentally drops her tool bag and it drifts away from the spacecraft.2 In this example,
P is the tool bag, O is the spacecraft, and O is fixed relative to the stars. Frame B
is attached to the spacecraft. We would like to understand the motion of the tool bag
relative to frame B, since this is the perspective of the astronaut. Suppose that during
the space walk the spacecraft thrusters are off, so IaO /O = 0, IαB = 0, and IωB is
constant. Furthermore, since there are no forces acting on the tool bag, IaP/O = 0.
Rearranging Eq. (10.52) gives
B
(10.53)
aP/O = −2IωB × BvP/O − IωB × IωB × rP/O .
Using the Cartesian coordinates (x, y, z)B to describe the position of the tool bag
relative to O and writing Eq. (10.53) in matrix notation with respect to frame B
yields
⎡
⎤
⎡ ⎤
⎡
⎤
⎡
⎤
⎡ ⎤
⎤
ω1
ẋ
ω1
ω1
x
ẍ
⎣ ÿ ⎦ = −2 ⎣ ω2 ⎦ × ⎣ ẏ ⎦ − ⎣ ω2 ⎦ × ⎣ ω2 ⎦ × ⎣ y ⎦ ,
ω3 B
ω3 B
ω3 B
ż B
z B
z̈ B
⎡
where ω1, ω2, and ω3 are the components of IωB expressed in frame B.
2 This
incident happened outside the International Space Station on November 18, 2008.
(10.54)
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CHAPTER TEN
20
z (m)
10
0
–10
–20
20
y( 0
m)
30
–20
10
0
x
20
(m)
Figure 10.16 Tool bag in space.
Now, without loss of generality, suppose that the (constant) angular velocity of the
spacecraft is aligned with the first axis of frame B, so that ω1 = 0, and ω2 = ω3 = 0.
Eq. (10.54) is equivalent to
ẍ = 0
ÿ = 2ω1ż + ω12y
z̈
(10.55)
= −2ω1ẏ + ω12z.
Integrating Eq. (10.55) using ω1 = 0.5 m/s, x(0) = y(0) = z(0) = 0 m, ẋ(0) = 0.3
m/s, ẏ(0) = 0.1 m/s, and ż(0) = 0.2 m/s yields the trajectory shown in Figure 10.16.
From the perspective of the spacecraft (small circle), the tool bag spirals away at a
constant rotational rate in the y-z plane and a constant (linear) rate in the x direction.
Example 10.14 The Translational Equations of Motion for an Airplane
This problem uses the results from Example 10.9 to find the translational equations of
motion for an airplane, that is, the dynamic equations that let us solve for the trajectory
of the plane in Cartesian coordinates in the inertial frame. Recall that we can use our
most basic results from the beginning of the book (and this chapter) to write three
simple equations of motion in Cartesian coordinates,
ẍ = fx /mG
ÿ = fy /mG
z̈ = fz /mG,
where fx , fy , and fz are the component magnitudes of the applied force in the inertial
frame. Note that here we have assumed that the separation principle for rigid bodies
PARTICLE KINEMATICS AND KINETICS IN THREE DIMENSIONS
445
still holds in three dimensions and that the airplane can be treated as a point mass
located at its center of mass, something not verified until the next chapter.
In the absence of any other information about the forces on the plane, these
equations seem like a reasonable approach. However, remember that we don’t usually
write the velocity of an airplane in terms of components in the inertial frame but rather
as components in the body frame,
I
vG/O = ub1 + vb2 + wb3.
It is thus convenient to have equations of motion for the body-frame components of
the velocity. This is easily done using the transport equation and Newton’s second
law applied to the center of mass:3
Bd FG
I
=
vG/O + IωB × IvG/O .
mG
dt
(10.56)
This equation looks a bit odd, as we are taking the derivative of the inertial velocity
relative to the body frame; we have never done that before. This approach is fine,
though. Remember, the transport equation works on any vector. Here we are just
applying it to the inertial velocity. This is a variation of the acceleration equation in
Eq. (10.52).
Performing the operations on the components of the vectors in Eq. (10.56) results
in the three scalar equations,
u̇ = rv − qw + F1/mG
(10.57)
v̇ = pw − ru + F2/mG
(10.58)
ẇ = qu − pv + F3/mG,
(10.59)
where F1, F2, and F3 are the three component magnitudes of the total force on the
plane in the body frame. Eqs. (10.57)–(10.59) are combined with the three kinematic
equations in Eq. (10.47) to get the six first-order equations of motion for airplane
translation. Of course, they can’t be integrated without explicit expressions for the
forces. We discuss the applied forces and the solution trajectories in Chapter 12.
10.5 Derivations—Euler’s Theorem
and the Angular Velocity
In this section we show that the angular velocity introduced above as an operator
corresponds to the instantaneous axis of rotation of body frame B in inertial frame
I . Note that the material in this section is not essential for the continuity of the rest
of the book. You can simply take our word that the angular velocity represents a
time-varying rotation about an instantaneous axis aligned with it—the so-called Euler
axis. However, we encourage you to read on, as a familiarity with Euler’s theorem is
extremely helpful in understanding the rotation of rigid bodies.
3 We
justify this approach in the next chapter.
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CHAPTER TEN
Recall that, when angular velocity was introduced in Chapter 3, it was obviously
aligned with the axis of rotation of the body. The first three parts of the book were
confined to considering planar motion and to body frames that rotated in absolute
space about a single axis only, usually denoted the 3-axis and represented by a unit
vector such as b3. The angular velocity was thus given by IωB = θ̇b3. We called this
a simple angular velocity. It was directed along the axis of rotation, and its magnitude
was given by the rate of rotation about that axis.
This chapter introduced the more general three-dimensional rotation of reference
frame B. We used our earlier result for the simple angular velocity, namely, that
it is an operator that provides unit-vector derivatives (by the transport equation) to
show that the transport equation still holds in three dimensions, where the angular
velocity is given by the sum of simple angular velocities associated with intermediate
frames, as in Eq. (10.44). We now show that this angular velocity is directed along the
instantaneous axis of rotation and that its magnitude is equal to the instantaneous rate
of rotation. To begin, we return to the relative orientation of two reference frames.
10.5.1 Euler’s Theorem
Section 10.4.1 showed that the relative orientation of frame B in I can be described
by three angles (the Euler angles) and that the magnitude of vector components can
be transformed from one frame to the other using the transformation matrix IC B . We
now introduce another way to describe the orientation of a rigid body (or reference
frame) relative to another frame: Euler’s Theorem of Rotation.
Theorem 10.1 Euler’s Theorem of Rotation The orientation of a reference frame
B relative to another frame I can be described by a single, simple rotation
of B, initially aligned with I , about an axis fixed in both B and I .
Figure 10.17 depicts the geometry of Euler’s theorem, where the two frames
I and B are initially aligned and then B rotates by an angle θ about the axis of
rotation L, whose direction is given by the unit vector k. Specifying k and the
rotation angle θ is another way of describing the orientation of B in I . Observe
that there are still three independent quantities describing the orientation (not four),
because k = k1e1 + k2e2 + k3e3 is a unit vector, thus implying the constraint k2 =
k12 + k22 + k32 = 1.
Recall also that the transformation matrix IC B always has one eigenvalue equal to
1 (as discussed in Section 10.4.1). The eigenvector associated with that eigenvalue is
the unit vector k! This is because if k is along the axis of rotation, it is necessarily the
same in both frames:
[k]I = IC B [k]B = [k]B .
This expression is an eigenvalue equation for matrix IC B with unity eigenvalue.
Remember that k is fixed in both I and B. Its components in I and B are the same
before and after the rotation, which implies
k = k1e1 + k2e2 + k3e3 = k1b1 + k2b2 + k3b3.
Euler’s theorem is a profound and important result that is useful for understanding
angular velocity. Before we prove Euler’s theorem, however, it is helpful to introduce
a corollary4 that is used both in our proof and in the discussion of angular velocity.
4 This
corollary follows the discussion of a simple rotation in Kane et al. (1983).
PARTICLE KINEMATICS AND KINETICS IN THREE DIMENSIONS
B, I
e3
a
k
L
k
θ
I
α2
L
θ
B
b3
b
b1
e2
α1
447
β2
b2
β1
e1
(a)
(b)
Figure 10.17 Change of vector a in absolute space to vector b after an Euler rotation of angle
θ about axis L. (a) Before rotation. (b) After rotation.
Corollary 10.1 (Kane et al. 1983) Consider an arbitrary vector a fixed in frame I
and a vector b fixed in B. If B is aligned with I prior to the rotation, such that
b = a, then after B undergoes a rotation by an amount θ about the line L (the
Euler axis), the vector b can be written in terms of a, the unit vector k along
the axis of rotation, and the angle θ by means of
b = a cos θ − (a × k) sin θ + (a . k)k(1 − cos θ).
(10.60)
To prove this corollary, we introduce two unit vectors fixed in I , α1 and α2, both
perpendicular to the rotation axis L, so that the three vectors α1, α2, and k form a righthanded orthogonal set, as in Figure 10.17a. Thus we can write vector a as components
in these unit-vector directions fixed in I ,
a = a1α1 + a2α2 + a3k.
(10.61)
We next introduce two unit vectors fixed in frame B, β1 and β2, also both perpendicular to L, and such that before the rotation α1 = β1 and α2 = β2. Likewise, the
three vectors β1, β2, and k form a right-handed orthogonal set fixed in B. Since b is
aligned with a before the rotation, its components in these directions are the same:
b = a1β1 + a2β2 + a3k.
(10.62)
Since b is fixed in B, these are its components after the rotation as well.
Now we express b as components in terms of α1, α2, and k after the rotation, as
shown in Figure 10.17b. This is done by writing the two unit vectors β1 and β2 in terms
of unit vectors α1 and α2. Since these unit vectors differ only by a simple rotation of
angle θ about a mutually orthogonal direction, k, we can use the transformation table
for a simple planar rotation to find
β1 = cos θα1 + sin θα2
β2 = − sin θα1 + cos θα2
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CHAPTER TEN
after the rotation. Substituting these into Eq. (10.62) gives
b = (a1 cos θ − a2 sin θ)α1 + (a1 sin θ + a2 cos θ)α2 + a3k.
This equation is the same result obtained by substituting the expression for a
in Eq. (10.61) into Eq. (10.60), thus verifying Corollary 10.1. Another nice check of
Eq. (10.60) is to substitute k for a and show that the result is b = k, as expected,
because k does not change with the rotation, as it is along the axis of rotation.
We can now use Eq. (10.60) to find expressions for the elements of the directioncosine matrix in terms of the component magnitudes of the unit vector k = k1e1 +
k2e2 + k3e3 and the rotation angle θ. Recall from Section 10.4.1 that the elements of
the direction-cosine matrix, I C B , between reference frame I = (O, e1, e2, e3) and
B = (O, b1, b2, b3) are given by
Cij = bi . ej
i, j = 1, 2, 3.
Using the fact that bi = ei prior to the rotation, we can substitute for bi from Eq.
(10.60) to obtain
Cij = ei cos θ − (ei × k) sin θ + (ei . k)k(1 − cos θ) . ej .
Using component magnitudes of k in I (or B) results in the following expressions
for each element of the transformation matrix:
C11 = cos θ + k12(1 − cos θ)
C12 = k3 sin θ + k1k2(1 − cos θ)
C13 = −k2 sin θ + k1k3(1 − cos θ)
C21 = −k3 sin θ + k1k2(1 − cos θ)
C22 = cos θ + k22(1 − cos θ)
C23 = k1 sin θ + k2k3(1 − cos θ)
C31 = k2 sin θ + k3k1(1 − cos θ)
C32 = −k1 sin θ + k3k2(1 − cos θ)
C33 = cos θ + k32(1 − cos θ).
As expected, we can write the transformation matrix either in terms of the three Euler
angles describing the orientation of B in I (as in Eq. (10.37)) or in terms of the
components of the unit vector k along the Euler axis of rotation and the rotation
angle θ. Either set is a valid description of the orientation of B in I .
There is something even more important about these relationships. We can use
them to prove Euler’s theorem. Suppose that the rigid body (reference frame B) has
some arbitrary orientation in I . This orientation could be described by the Euler angles
(ψ, θ, φ)IB , as in the previous section. And as in Section 10.4.1, we can write the
transformation matrix in terms of these angles. The inverse of the above relationships
then provides expressions for the components of the unit vector k along the Euler axis
of rotation and the rotation angle in terms of the components of this matrix:
PARTICLE KINEMATICS AND KINETICS IN THREE DIMENSIONS
θ = arccos
C11 + C22 + C33 − 1
2
449
(10.63)
C23 − C32
2 sin θ
C − C13
k2 = 31
2 sin θ
C − C21
k3 = 12
.
2 sin θ
k1 =
(10.64)
(10.65)
(10.66)
Eqs. (10.63)–(10.66) constitute a proof of Euler’s theorem! In other words, given
any arbitrary orientation and the transformation matrix (or Euler angles) describing it,
we can always find a single axis of rotation and an angle about that axis corresponding
to the orientation by using Eqs. (10.63)–(10.66).
10.5.2 The Angular Velocity
Finally, we turn our attention back to the angular velocity. Suppose now that the rigid
body (frame B) is changing orientation with time in I . The rate of change of vectors
is given by the transport equation, but what does the angular velocity physically
represent? If the orientation of B is changing with time, then over any small time
interval, that change could be represented by an Euler axis rotation by using Euler’s
theorem. Thus the Euler axis represents the instantaneous axis of rotation. If we again
consider two vectors a and b, with a fixed in I and b fixed in B and initially aligned,
then a very short time later the vector b is given by Eq. (10.60) in Corollary 10.1 (where
θ is small). What, then, is the rate of change of b? Taking the inertial derivative of
Eq. (10.60) yields
Id
dt
b = −θ̇ (a sin θ + (a × k) cos θ − (a . k)k sin θ),
(10.67)
where k is along the instantaneous axis of rotation and we have used the fact that a
and k are fixed in I . We also know the derivative of b from the transport equation:
Id
dt
b=
Bd
b +IωB × b.
dt
(10.68)
=0
Eqs. (10.67) and (10.68) for the inertial velocity of b should, of course, be the same.
Let us suppose that IωB = θ̇k, that is, that the angular velocity is a simple rotation of
rate θ̇ about the Euler axis. Then substituting this expression for IωB into Eq. (10.68)
and using the expression for b in Eq. (10.60) gives
I B
ω × b = θ̇[(k × a) cos θ − k × (a × k) sin θ + (a . k) k
× k(1 − cos θ)]
= −θ̇ (a × k) cos θ + a sin θ − (a . k)k sin θ ,
=0
(10.69)
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CHAPTER TEN
where we have used the triple vector cross product identity from Appendix B: k ×
(a × k) = a(k . k) − k(a . k). Eq. (10.69) is identical to Eq. (10.67), as required.
This shows that the angular velocity is indeed given by IωB = θ̇k; it is directed along
the instantaneous Euler axis and is equal in magnitude to the instantaneous rate of
rotation about that axis. The angular velocity represents the rotation rate about an
instantaneous axis of rotation parallel to it.
10.6 Tutorials
Tutorial 10.1 The Local Vertical
Example 10.2 introduced the geocentric inertial frame I = (O, e1, e2, e3), and the
geographic frame G = (O, g1, g2, g3); the former is an inertial frame located at the
center of the earth and the latter is located at the earth’s center and fixed to the earth.
The earth rotates in absolute space at a rate E (i.e., once per day). The angular
velocity is
I G
ω =
E g3.
In this tutorial we are interested in examining the equations of motion of particles
relative to a “local” frame on the surface of the earth. That is, we consider a point O
somewhere on the surface of the earth. Figure 10.18a locates that point with the spherical coordinates (RE , θ, φ)G, where RE is the earth’s radius, θ is the longitude, and φ
is the co-latitude (90◦ minus the latitude). We then place frame B = (O , b1, b2, b3)
at O with b3 along the radius direction to the center of the earth, b1 pointed south
(i.e., perpendicular to b3 and in the plane formed by b3 and g3), and b2 = b3 × b1 (as
shown in Figure 10.18b).
G
ΩE
ϕ
O′
b2
b1
O
θ
g1
(a) Coordinates
B
O′
rO′/O
O
b3
g3
g2
(b) Reference frames
Figure 10.18 A spherical earth with point O located on the surface by spherical coordinates
φ and θ. Reference frame G is located at the center O of the earth fixed to the earth’s surface
and rotating in absolute space with angular velocity IωG = E g3. Local frame B is also fixed
to the earth at O .
PARTICLE KINEMATICS AND KINETICS IN THREE DIMENSIONS
451
Because the earth is very large, for most problems considered on the surface of
the earth, B can be treated as inertial (this assumption holds for almost every problem
we have considered in the book). For problems involving gravity near the surface of
the earth, for instance, we usually assume that the force of gravity is constant and
is always directed downward in frame B (i.e., in the b3 direction). This tutorial tests
how good an assumption that is.
To consider the dynamics in frame B of particle P , located at rP/O , we begin with
the expression for the inertial acceleration of P in Eq. (10.52) and Newton’s second
law,
aP/O = IaO /O + BaP/O + 2IωB × BvP/O + IωB ×
I
F
ω × rP/O = P ,
mP
I B
(10.70)
where the angular acceleration has been set to zero. (Because the local frame B is
fixed to the earth, it has zero angular velocity relative to the geographic frame G and
thus IωB = IωG .)
We are interested in finding equations of motion for P in the local frame. We
rearrange Eq. (10.70) slightly to solve for the acceleration of P in B:
B
aP/O =
FP
− IaO /O − 2IωG × BvP/O − IωG × IωG × rP/O .
mP
(10.71)
We immediately see that P behaves slightly differently than we would have expected
had the local frame been inertial. Its motion is affected by the Coriolis and centripetal
terms as well as the acceleration of the origin O of the frame. To complete the
description, we need to solve for that acceleration. That is done using the transport
equation, as in Chapter 8. The inertial velocity of O is
I
vO /O =
rO /O +IωG × rO /O ,
dt Gd
=0
where the derivative in the G frame is zero because point O is fixed to the surface of
the earth. Taking the inertial derivative again gives the acceleration of O :
aO /O = IωG × (IωG × rO /O ).
I
Next we find the components of this acceleration in the geographic frame. Using
spherical coordinates for the position (Eq. (10.4)), we have
rO /O = RE cos θ sin φg1 + RE sin θ sin φg2 + RE cos φg3.
The acceleration of O is
I
aO /O = −RE 2E (cos θ sin φg1 + sin θ sin φg2).
Finally, it is more convenient to express the acceleration of O in terms of components in the local frame, as that is where we want to find equations of motion of P .
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CHAPTER TEN
Because the local frame B is the same as a spherical frame, we use the transformation
table in Eq. (10.13):
b1
b2
b3
g1
g2
g3
cos θ cos φ
− sin θ
cos θ sin φ
sin θ cos φ
cos θ
sin θ sin φ
− sin φ
0
cos φ .
Transforming the acceleration gives (after a bit of algebra)
I
aO /O = −RE 2E (cos φ sin φb1 + sin2 φb3).
(10.72)
It should not come as a surprise that this acceleration is independent of longitude due
to the rotational symmetry about g3.
We now use Eq. (10.71) and the acceleration in Eq. (10.72) to solve the following
problem. We consider a particle P at rest at the origin of the local frame. Because
gravity is acting, there must be a reaction force to keep the particle stationary (applied
by the earth’s surface). Because the particle is at the origin and at rest, rP/O and BvP/O are both zero, leaving us, from Newton’s second law, with
0 = −gb3 +
N
+ RE 2E (cos φ sin φb1 + sin2 φb3),
mP
where N is the reaction force from the ground. Solving for N yields
N = mP (g − RE 2E sin2 φ)b3 − mP RE 2E cos φ sin φb1.
N is not vertical! That is, the reaction force keeping the particle stationary is not
directed along the line through the center of the earth, but is offset due to the rotation
of the earth. If we were to drop an object, it would not fall toward the center of the
earth but rather along the direction −N. We call this direction the local vertical. For
instance, if we were to hang a pendulum, its equilibrium would be along the local
vertical rather than the geographic one. A pendulum that identifies the local vertical
is called a plumb bob.In fact, there is no local measurement we can take to differentiate
between the local gravity vector and the one due only to the earth’s mass.
How big is this effect? If we define the small unitless quantity by
=
RE 2E
g
,
then we can find the angle between N and the b3 direction by taking their dot product
and dividing by N. Letting β be the angle between the local vertical and b3, we
have
cos β =
1 − sin2 φ
.
(10.73)
2 sin2 φ cos2 φ + (1 − sin2 φ)2
The radius of the earth is approximately 6,378 km, and its angular rotation rate is
roughly 7.3 × 10−5 rad/s. Assuming the acceleration due to gravity at the surface is
g = 9.8 m/s2, we find that = 3.4 × 10−3. Suppose we consider the local vertical at
PARTICLE KINEMATICS AND KINETICS IN THREE DIMENSIONS
453
Figure 10.19 Watt flyball governor. Drawing from Routledge (1900).
a latitude of φ = 45◦ (somewhere in Canada, say). Then, from Eq. (10.73), it is offset
from the earth’s radial direction by an angle of roughly 0.1◦.
As a final note, Eq. (10.71) can be used to estimate the error made when assuming
that a frame fixed to the surface of the earth is inertial. In most of our examples, we
have treated earth-fixed frames as inertial. That is only an approximation, as the earth
is rotating. The three terms in Eq. (10.71) provide the correction to account for the
earth’s rotation.
Tutorial 10.2 The Flyball Governor
The flyball governor is one of the earliest examples of an automatic control system.
Its original purpose was to regulate the speed of the grinding stone in a flour mill.
However, it was made famous by James Watt in the late eighteenth century when he
adapted it to maintain the speed of a steam engine under load. An illustration of the
Watt flyball governor is shown in Figure 10.19. The vertical motion of the balls opens
or closes a steam valve that is connected to a rotary engine, the spin rate of which
determines the motion of the balls.
This tutorial derives the equation of motion of a simple flyball-governor model
using point masses and massless rods. Assume > 0 is the rotational speed of the
shaft (D in Figure 10.19) and ignore the lever control system. We find that the angle
the balls make with the rotating shaft is determined by the shaft rotation rate. This
is an excellent example of solving a particle-dynamics problem in three dimensions
with multiple rotating frames.
Our model of the governor is shown in Figure 10.20. There are three masses: the
two balls on the swinging massless rods (they are simple pendula) and the bottom
mass that moves up and down the shaft. The position of the lower mass M is given
by y and the angle of the rods holding the balls is given by θ. Nevertheless, this
454
CHAPTER TEN
B b2 C
c2
O
θ
I e2
b1
e1
c1
l
y
m
–T1c1
g
T2
–mgb2
l
M
T2
Ω
Ω
(a) Coordinates
T2
–Mgb2
(b) Reference frames
(c) Free-body diagrams
Figure 10.20 Simple model of the flyball governor.
is a single-degree-of-freedom problem because of the linked rods: the geometric
constraint between y and θ is
y = 2l cos θ.
(10.74)
By symmetry we need only find the equation of motion for one of the balls; the other
ball, because of the constraint, behaves exactly the same way.
We introduce three frames as shown in Figure 10.20b. As usual, I = (O, e1, e2, e3)
is an inertial frame fixed in absolute space. Frame B = (O, b1, b2, b3) is an intermediate frame rotating with the shaft at IωB = b2. Frame C = (O, c1, c2, c3) is a
body frame fixed to the pendulum rod and rotating in frame B with angular velocity BωC = θ̇b3. We use the addition property of angular velocities to find the angular
velocity of the body frame C in I :
ω = IωB + BωC
I C
= b2 + θ̇b3.
Frame C is related to frame B by the transformation table
c1
c2
c3
b1
b2 ,
sin θ
cos θ
0
− cos θ
sin θ
0
b3
0
0
1 .
We proceed by finding the equation of motion for θ, using the free-body diagrams
in Figure 10.20c. The kinematics of the lower mass are
rM/O = −yb2
I
vM/O = −ẏb2
I
aM/O = −ÿb2.
From the free-body diagram, the force on M is
FM = (2T2 cos θ − Mg)b2.
PARTICLE KINEMATICS AND KINETICS IN THREE DIMENSIONS
455
Newton’s second law then gives
−M ÿ = 2T2 cos θ − Mg.
Solving for T2 yields
T2 =
M
(g − ÿ) .
2 cos θ
(10.75)
We can eliminate ÿ by differentiating Eq. (10.74) twice to obtain
ÿ = −2l(θ̇ 2 cos θ + θ̈ sin θ).
(10.76)
Substituting Eq. (10.76) into Eq. (10.75) yields
T2 =
Mg
+ Ml θ̇ 2 + Ml θ̈ tan θ.
2 cos θ
(10.77)
The equation of motion for the ball is found by using angular momentum, which
avoids having to solve for the internal force T1. The position expressed in the body
frame is
rm/O = lc1.
The velocity is found from the transport equation to be
vm/O = IωC × lc1
I
= (b2 + θ̇b3) × l(sin θb1 − cos θb2)
= θ̇l cos θb1 + θ̇l sin θ b2 − l sin θ b3.
We use this expression to compute the angular momentum of a ball in absolute space:
hm/O = rm/O × mIvm/O
I
= l(sin θb1 − cos θb2) × m(θ̇l cos θb1 + θ̇l sin θ b2 − l sin θ b3)
= ml 2 sin θ cos θ b1 + ml 2 sin2 θb2 + mθ̇l 2b3.
(10.78)
The equation of motion comes from setting the inertial derivative of the angular
momentum equal to the moment about O. The moment is found from the forces on
m in the free-body diagram:
Mm/O = rm/O × Fm
= l(sin θb1 − cos θb2)
× (−T1(sin θb1 − cos θb2) − T2(sin θb1 + cos θb2) − mgb2)
= (−2T2l sin θ cos θ − mgl sin θ)b3.
Perhaps not surprisingly, the moment on the ball is only in the b3 direction,
implying that the components of the angular momentum in the b1 and b2 directions
are conserved. Thus, to find the equation of motion for the ball, we use the transport
456
CHAPTER TEN
equation on the angular momentum and set the b3 component equal to the moment.
Taking the inertial derivative of Eq. (10.78) gives
Id
dt
(Ihm/O ) =
Bd
dt
(Ihm/O ) + IωB × Ihm/O
= mθ̇l 2(cos2 θ − sin2 θ)b1 + 2mθ̇l 2 sin θ cos θ b2 + mθ̈l 2b3
+ b2 × m(l 2 sin θ cos θb1 + l 2 sin2 θb2 + θ̇l 2b3).
Setting the b3 component equal to the moment yields the second scalar equation of
motion:
ml 2(θ̈ − 2 sin θ cos θ) = −2T2l sin θ cos θ − mgl sin θ.
Our final step is to substitute for T2 from Eq. (10.77) and solve for θ̈ to yield our
desired equation of motion for the flyball governor:
θ̈ =
−(m + M)g + (m2 − 2M θ̇ 2)l cos θ sin θ
(m + 2M sin2 θ)l
.
The equation of motion is often simplified for the case M m by dividing through
by m and dropping terms of order M
m 1. The approximate equation of motion is
g
2
θ̈ ≈ − + cos θ sin θ.
l
Our interest in the flyball governor is motivated by its equilibrium solutions. Since
it is used as a feedback controller, we are interested in seeing whether there is a steadystate solution with nonzero θ that keeps fixed. Setting θ̈ = 0 gives the following
condition for an equilibrium:
g
2
− + cos θ sin θ = 0.
l
The equilibrium condition is satisfied by the equilibrium angles θ1∗ = 0, which
correspond to the balls hanging vertically downward. There is also an additional pair
of equilibrium points at
g
θ2∗ = ± arccos
,
l2
as long as
g
≤ 1.
l2
The latter condition is satisfied for ≥ 0, where
g
0 =
.
l
(10.79)
PARTICLE KINEMATICS AND KINETICS IN THREE DIMENSIONS
457
2
80
θ (Hz)
θ (°)
60
40
0
20
0
0
2
Ω (Hz)
–2
4
(a)
–50
0
θ (°)
50
(b)
θ (Hz)
2
0
–2
–50
0
θ (°)
50
(c)
Figure 10.21 Flyball governor with M m and l = 0.1 m. (a) Bifurcation diagram showing
bifurcation value 0 = 1.58 Hz. (b) Phase portrait for < 0 showing one equilibrium plotted
as a dot. (c) Phase portrait for > 0 showing three equilibria.
This analysis shows that the flyball governor exhibits a fascinating behavior known
as a bifurcation. Depending on the value of , the system has either one equilibrium
point or three equilibrium points. We call the bifurcation parameter, and the critical
value where the number of equilibrium points changes is called the bifurcation value.
In this case, the bifurcation value is given by Eq. (10.79).
We can graphically represent this behavior in a bifurcation diagram, a plot of the
equilibrium values as a function of the bifurcation parameter. Such a diagram for the
flyball governor is given in Figure 10.21a. We can also often graph the trajectories
of the system in a phase portrait. In a phase portrait, rather than plotting the state
variables versus time (here, θ and θ̇), we instead plot them versus each other for
various trajectories. This plot gives a picture of the overall behavior of the system for
different initial conditions. For example, the phase portrait of a spring-mass system
consists of just circles of varying sizes. An isolated dot on the phase portrait is an
equilibrium solution: a location where the system remains for all time. Phase portraits
are excellent tools for visually displaying equilibria. Figure 10.21b shows a phase
portrait for the flyball governor for < 0. Note that there is only one equilibrium at
θ = 0 (represented by the dot). Figure 10.21c shows a phase portrait of the flyball
governor for > 0. Note that the equilibria have
into three possible
bifurcated
g
∗
∗
values, corresponding to θ1 = 0 and θ2 = ± arccos l2 .
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CHAPTER TEN
10.7 Key Ideas
.
This chapter introduced three three-dimensional coordinate systems: Cartesian
coordinates, (x, y, z)I , cylindrical coordinates, (r, θ, z)I , and spherical coordinates, (r, θ, φ)I . The position of P with respect to O expressed in terms of
vector components in frame I = (O, ex , ey , ez ) using each of these coordinate
systems is
rP/O = xex + yey + zez
= r cos θex + r sin θey + zez
= r cos θ sin φex + r sin θ sin φey + r cos φez
.
(Cartesian)
(cylindrical)
(spherical).
The cylindrical frame Bc = (O, er , eθ , ez ) is like a polar frame, except that the
unit vector er points at the projection of P onto the x-y plane, instead of at P
directly. The kinematics of P using cylindrical coordinates in the cylindrical frame
are
rP/O = rer + zez
I
vP/O = ṙer + r θ̇ eθ + żez
I
aP/O = (r̈ − r θ̇ 2)er + (2ṙ θ̇ + r θ̈)eθ + z̈ez .
.
The spherical frame Bs = (O, eφ , eθ , er ) is defined with respect to an inertial
frame I using two simple rotations and one intermediate frame. The unit vector
er is directed from O to P . The kinematics of P using spherical coordinates in the
spherical frame are
rP/O = rer
I
vP/O = r φ̇eφ + r θ̇ sin φeθ + ṙer
I
aP/O = (2ṙ φ̇ + r φ̈ − r θ̇ 2 cos φ sin φ)eφ
+ (2ṙ θ̇ sin φ + 2r θ̇ φ̇ cos φ + r θ̈ sin φ)eθ
+ (r̈ − r φ̇ 2 − r θ̇ 2 sin2 φ)er .
.
.
.
The expressions for three-dimensional linear momentum, angular momentum,
and energy of a particle are identical to the two-dimensional expressions.
The Euler angles (ψ, θ, φ)IB are a set of coordinates describing the orientation of
a three-dimensional body frame B with respect to frame I . A set of Euler angles
is defined by three simple rotations and two intermediate frames.
The entries in the three-dimensional transformation table that relates the unit
vectors of frame B to frame I form a 3 × 3 transformation matrix IC B that is
used to transform the components of a vector from B to I or vice versa. In matrix
notation we have
[rP/O ]I = IC B [rP/O ]B .
PARTICLE KINEMATICS AND KINETICS IN THREE DIMENSIONS
.
459
The transport equation is valid in three dimensions:
Ad
dt
a=
Bd
dt
a + AωB × a,
where a is an arbitrary vector, and AωB is the angular velocity of frame B with
respect to frame A.
.
.
The angular-velocity addition property holds in three dimensions.
The three-dimensional kinematics of a particle in a rotating frame B with origin
O are
vP/O = IvO /O + BvP/O + IωB × rP/O I
aP/O = IaO /O + BaP/O + IαB × rP/O + 2IωB × BvP/O + IωB × IωB × rP/O ,
I
where IαB =
.
Bd
dt (
IωB ) = Id IωB
dt
is the angular acceleration.
The angular velocity in terms of the rates of change of the 3-2-3 Euler angles
(ψ, θ, φ)IB is
I B
ω = ψ̇a3 + θ̇c2 + φ̇b3,
which yields the kinematic equations of rotation
ω1 = θ̇ sin φ − ψ̇ sin θ cos φ
ω2 = θ̇ cos φ + ψ̇ sin θ sin φ
ω3 = φ̇ + ψ̇ cos θ.
.
.
The angular velocity represents the rate of rotation about an instantaneous axis of
rotation aligned with it.
Euler’s theorem of rotation states that any change in the orientation of frame B
in frame I can be produced by a single, simple rotation about an axis fixed in both
B and I .
10.8 Notes and Further Reading
Our primary reference for much of the material on the kinematics of rotation in this
chapter is the book on spacecraft dynamics by Kane et al. (1983). This book has
an excellent and thorough treatment of several different descriptions of orientation
(including quaternions, Rodriguez parameters, and both body and space angles) as
well as the kinematics of rigid bodies. It is an excellent source for delving further. It
is also the only text we know of that discusses all of the different sets of Euler angles,
including two appendices with every direction-cosine matrix and corresponding set
of kinematic equations of rotation. Although all textbooks discuss Euler angles, not
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CHAPTER TEN
all use the same set, and some may switch for different applications (e.g., the standard
set used for spacecraft dynamics is different from that used for airplane dynamics).
Our discussion of the physical interpretation of the velocity and Euler’s theorem
is also motivated by Kane’s books (Kane 1978; Kane and Levinson 1985; Kane et al.
1983). Many introductory texts, in both physics and engineering, use infinitesimal
rotations and qualitative arguments to explain the angular velocity. We prefer to treat
it as an operator and show, using Euler’s theorem, that it is along the instantaneous
axis of rotation. Eq. (10.60) in Corollary 10.1 and its derivation come directly from
Kane et al. (1983), as does our statement of Euler’s theorem.
10.9 Problems
10.1 Write the position of P with respect to O in frame I as shown in Figure 10.22
using
a. Cartesian coordinates,
b. cylindrical coordinates,
c. spherical coordinates.
I
ez
ey
2.27 m
3.64 m
O
1.92 m
P
4.71 m
1.23 m
ex
Figure 10.22 Problem 10.1.
10.2 Let I = (O, e1, e2, e3) be an inertial reference frame, A = (O, a1, a2, a3) be
a cylindrical frame, and B = (O, eφ , eθ , er ) be a spherical frame, as shown in
Figure 10.23. Derive the following:
a. The transformation tables between A and I , and between B and A.
b. The angular velocity of A with respect to I , and of B with respect to
I.
c. The position rP/O and velocity IvP/O of P with respect to O, expressed
as components in A using cylindrical coordinates (r, θ, z)I .
d. The position rP/O and velocity IvP/O of P with respect to O, expressed
as components in B using spherical coordinates (r, θ, φ)I .
PARTICLE KINEMATICS AND KINETICS IN THREE DIMENSIONS
I
A
a3
e3, a3
P
ϕ
a2
P
er
O
a 2 , eθ
O
e2
θ
e1
461
eϕ
a1
a1
B
Figure 10.23 Problem 10.2.
10.3 Let C = (O, eφ , eθ , er ) be a spherical frame and I = (O, e1, e2, e3) be an
inertial frame such that IωC = −θ̇ sin φeφ + φ̇eθ + θ̇ cos φer . Derive
and
Id
Id
dt (eθ )
dt (er ).
10.4 Solve for the equation of motion of the spherical-pendulum angle φ using
angular momentum. Assume it rotates at a constant angular speed . See
Figure 10.24.
O
g
ϕ
P
Ω
Figure 10.24 Problem 10.4.
10.5 Find the angular velocity of a football spinning at about its long axis and
traveling at a constant speed v0 tangent to a semicircle of radius R, as shown
in Figure 10.25.
v0
Ω
R
Figure 10.25 Problem 10.5.
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CHAPTER TEN
10.6 Repeat Problem 10.5 where the football travels along a parabolic trajectory.
10.7 A particle of mass m slides without friction inside a three-dimensional conical funnel, whose walls have unit slope and whose center axis is vertical
(Figure 10.26). If the particle is launched inside the funnel at height h and
horizontal speed s0, find the vertical speed of the particle when it reaches the
bottom of the funnel. [HINT: Use a cylindrical frame and cylindrical coordinates to write the kinematics of P with respect to O.]
I
P
g
1
h
O
Figure 10.26 Problem 10.7.
10.8 Find the conditions for the steady-state solution of a spherical pendulum at a
nonzero angle other than π.
10.9 Integrate the equations of motion for the flyball governor for different and
show the bifurcation discussed in Tutorial 10.2.
10.10 Example 10.3 derived the equations of motion for a spherical pendulum. When
dealing with a real pendulum, however, we are actually operating in a rotating
frame. Repeat the derivation of the equations of motion, but this time, place the
pendulum in a local vertical reference frame at latitude β (see Tutorial 10.1).
You may use whatever reference frames and coordinate systems you consider
to be appropriate, but remember that your equations of motion must reflect the
motion of the pendulum in all three dimensions relative to the local vertical.
a. Using your new equations of motion, explain what effect the rotation
of the earth has on the plane of the pendulum’s swing.
b. French physicist Jean-Bernard-Léon Foucault (1819–1868) predicted
that a pendulum could be used to prove the rotation of the earth. In
1851, a 67 m pendulum with a 28 kg bob was constructed in Paris
(latitude 48◦ N). Estimate the rate of rotation of the plane of this
pendulum’s swing.
c. What happens if a Foucault pendulum is constructed at the north (or
south) pole? At the equator?
10.11 Suppose you are operating an unmanned surveillance aircraft (UAV, or unmanned aerial vehicle) from a ground station at point O, as shown in Figure 10.27. Your job is to search for and take a picture of a target ground facility
over the next ridge. Assume a flat earth. Suppose the UAV flies level at constant speed v0 (its velocity always points through its nose, and its wings are
PARTICLE KINEMATICS AND KINETICS IN THREE DIMENSIONS
v0
O
l
463
P
O′
A
Rc
T
Figure 10.27 Bird’s-eye view of system in Problem 10.11. Point P represents the UAV. Point
T represents the target facility, but the target is not necessarily located as shown in the figure.
parallel to the ground), counter-clockwise in a circle of radius RC centered at
a surveyed point O a distance l away from your station and at altitude h. Let
B be a frame fixed to the aircraft and C be a frame fixed to a camera mounted
under the aircraft, which can rotate with respect to B on gimbals. The camera
lens points in the c3 direction.
a. Suppose the camera sees the facility when the UAV is at a position on
its circular trajectory on the line between O and O (point A in Figure
10.27). At the time, the orientation of the camera with respect to the
UAV is given by the 3-2-3 Euler angle set (ψ = ψA, θ = θA, φ = 0)BC ,
and the radar ranging gives its distance along the line of sight to the
camera as d = dA m. What is the position of the target from your station
in Cartesian coordinates (useful for future targeting)?
b. The UAV now must keep the target in the camera’s sights as it continues to fly in its circle. Find expressions for the rate of change of the
Euler angles so that the camera stays pointed at the target. Assume that
this can be done with φ kept at zero for all time (since the camera is
symmetric about the axis through the lens). Find a system of differential equations that would allow you to solve for ψ, θ, and d for all
time.
10.12 Consider an experimental high-altitude vehicle dropped from a large C-130
transport plane. The plane and vehicle are beyond the visible range of the
ground system. Sensors on the plane measure the velocity IvP/O of the released vehicle relative to the C-130. The C-130 telemeters this information
to the ground observers as well as its own inertial velocity IvO /O , its angular
velocity in inertial space IωB , and its inertial position rO /O , obtained from its
inertial navigation system and GPS sensors. The ground observers must point
their cameras and other equipment to pick up the experimental vehicle; thus
they must know the vehicle’s velocity in inertial space. Write down the vector
expression they use to find the vehicle’s velocity from the telemetered data.
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10.13 Consider the spinning disk in a single gimbal shown in Figure 10.28. The disk
has a diameter of 50 mm, and the gimbal has a diameter of 100 mm. Find the
inertial velocities of points A and B if the rotor is spinning at 10 rad/s while
the gimbal rotates about the vertical axis at 2 rad/s. Express your answer as
components in an intermediate frame fixed to the gimbal but not spinning with
the rotor.
2 rad/s
50 mm
B
A
10 rad/s
100 mm
Figure 10.28 Problem 10.13.
CHAPTER ELEVEN
Multiparticle and Rigid-Body Dynamics in
Three Dimensions
This chapter concludes our study of Newtonian mechanics with an examination of
the motion of multiparticle systems and rigid bodies in three dimensions. By the end
of the chapter you will have all the tools and skills you need to tackle any dynamics
problem. Although it may seem like a fairly minor extension to move from the simple
planar rotations of Chapter 9 to the general three-dimensional rotations studied in
this chapter, you will see that a rich set of behavior emerges. The complex motion of
spinning bodies has allowed for an amazing array of engineered devices from toys to
spacecraft.
11.1 Euler’s Laws in Three Dimensions
We begin this chapter with the simple observation that the basic ideas of Chapters 6
and 7 translate unchanged to three dimensions. We were careful in those chapters to
develop our ideas using a vector treatment so that the fundamental results were independent of dimension. Thus the law of conservation of total linear momentum is
the same in three dimensions, and Definition 6.2 of the center of mass still applies,
including the center-of-mass corollary in Eq. (6.15). The discussion of collisions extends to three dimensions because we considered two objects whose impact velocities
are merely confined to a plane, as is the case in three dimensions in the absence of
external forces. The idea that a multiparticle system can be treated by applying Newton’s second law to each particle still holds—except we now have three coordinates
for each particle rather than two. Chapter 10 covered the three-dimensional equations
of motion of a particle.
Also still true is the all-important result from Chapter 6—that the center of mass of
a particle collection follows a trajectory determined by solving Newton’s second law
independently of the relative motion of the particles. That is, the three-dimensional
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CHAPTER ELEVEN
B
b3
h
R
b2
b1
Figure 11.1 Computing the center of mass of a continuous rigid body.
translational dynamics of the center of mass of a collection of particles is
(ext)
FG
= mGIaG/O ,
(11.1)
(ext)
represents the total external force acting on the system. When applied to
where FG
a rigid body, Eq. (11.1) is equivalent to Euler’s first law (Law 9.1) in three dimensions.
Example 11.1 Computing the Center of Mass of a Continuous Rigid Body
in Three Dimensions
Example 9.1 computed the center of mass of a uniform planar triangle. The same
equations allow us to find the center of mass of a continuous three-dimensional
body B. Consider the cone in Figure 11.1, which has base radius R and height
h. Because the cone can be easily described as a surface of revolution about the
b3 axis, it makes sense to use cylindrical coordinates here. As before, we start by
writing the position rdm/O = r cos θb1 + r sin θb2 + zb3 and the volume element
dV = rdrdθdz. We also assume a mass m and uniform density ρ. We apply Eq. (9.2)
to write the center of mass as
ρ
rG/O =
(r cos θb1 + r sin θb2 + zb3)rdrdθdz.
m B
To find the limits of integration for the three integrals, we must consider the shape
of the cone. The limits on θ and z are simple—θ goes from 0 to 2π, and z goes
from 0 to h. The upper limit on r, however, depends on the slope of the cone and is
MULTIPARTICLE AND RIGID-BODY DYNAMICS IN THREE DIMENSIONS
467
given by R(h − z)/ h. Finally, before performing the integration, we note that, if we
integrate over θ first, the b1 and b2 terms go to zero, thereby simplifying the remaining
integrations:
ρ
rG/O =
m
ρ
=
m
=
ρ
m
h
0
0
0
h
0
R(h−z)/ h
0
(r cos θb1 + r sin θb2 + zb3)rdθdrdz
R(h−z)/ h
2πrzb3drdz
0
h
2π
πR 2z(h − z)2
b3dz
h2
ρ h2πR 2
b3.
m 12
Because m = p B dV = ρπR 2h/3, the center-of-mass position is
=
rG/O =
h
b3.
4
This result is very useful because it states that the center of mass of any right circular
cone will be one-fourth of the way up its central axis.
Our treatment of the angular momentum of multiparticle systems and rigid bodies also extends to three dimensions. Recall that the total angular momentum of a
collection of particles (Definition 7.1) is
I
hO =
N
I
hi/O =
i=1
N
mi ri/O × Ivi/O .
i=1
For a continuous rigid body,
I
hO =
B
rdm/O × Ivdm/O dm.
When the internal-moment assumption holds, the angular-momentum form of
Newton’s second law in Eq. (7.5) applies for the total angular momentum of the
collection:
Id
dt
N
(ext)
.
hO =
ri/O × Fi(ext) = MO
I
(11.2)
i=1
These equations are still true in three dimensions, so that the derivations in Section 7.1 still hold. For rigid bodies, everything in Section 9.1 through Section 9.2 and
Section 9.3.1 applies unchanged in three dimensions. We did not use the planar assumption until Section 9.3.2. Thus the internal-moment assumption still applies as a
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CHAPTER ELEVEN
property of a rigid body and Euler’s second law in Eq. (9.6) holds unchanged for the
motion of a rigid body relative to an inertially fixed point O:
Id
hO = M O ,
I
dt
where we have dropped the superscript (ext) for a rigid body.
The other important result from Chapter 7 is the separation principle in Section 7.2:
the total angular momentum of the collection separates into the angular momentum
of the center of mass G relative to O and the angular momentum about the center of
mass. Those results were completely general and, consequently, still apply in three
dimensions. We have
I
hO = IhG/O + IhG,
where
I
hG/O = mGrG/O × IvG/O
and
I
hG =
N
mi ri/G × Ivi/G.
i=1
For a continuous rigid body,
I
hG =
B
rdm/G × Ivdm/G dm.
Again, the center of mass obeys Newton’s second law (Euler’s first law) as if the
collection were a single particle of mass mG located at G. When the internal-moment
assumption holds, the angular momentum hG about the center of mass satisfies the
equation of motion
Id
dt
I
hG =
N
(ext)
ri/G × Fi(ext) = MG
.
i=1
For a rigid body, the internal-moment assumption always holds, and this equation is
Euler’s second law about the center of mass:
Id I
hG = M G .
(11.3)
dt
Even before developing three-dimensional expressions for the angular momentum
in terms of the angular velocity and mass properties of a rigid body, we can already
solve interesting and useful problems just by using the angular-momentum expressions in Euler’s first and second laws. This observation is very important; in many
MULTIPARTICLE AND RIGID-BODY DYNAMICS IN THREE DIMENSIONS
469
Figure 11.2 Toy gyroscope on its stand. Image courtesy of Shutterstock.
problems it is useful to go back to these basic principles rather than jump right to the
more sophisticated tools you will soon learn. You will be surprised how much can be
done just by examining angular momentum.
Example 11.2 The Gyropendulum
This example derives the equations of motion for an inverted spherical gyropendulum.
The device is very similar to the spherical pendulum studied in Example 10.3 except
that, instead of the mass particle at the end of the massless rod, there is a spinning
rotor. This is an excellent model of the toy gyroscope depicted in Figure 11.2; when
spinning, the gyro doesn’t tip over but precesses around the vertical. The result derived
here qualitatively predicts this motion.
Our simple model of the gyropendulum, including the reference frames and freebody diagram, is shown in Figure 11.3. If the rotor radius is small compared to the
length of the pendulum, we can very effectively model it as a single particle with a
constant angular momentum of magnitude h about its symmetry axis. This approximation is common in three-dimensional dynamics, when the motion is dominated by
the effect of the large-spin angular momentum.
To begin, we review the results of Example 10.3 for the nonspinning case. We
still locate P using spherical coordinates (Figure 11.3a) and introduce the spherical
frame Bs (Figure 11.3b). The only difference here is that we are considering motion for
θ
er
er
mP
I
ϕ l
BS
–Fer
ez
g
eθ
ex
–mPgez
ey
eϕ
(a) Coordinates
(b) Reference frames
(c) Free-body diagram
Figure 11.3 Spherical gyropendulum using spherical coordinates and the spherical frame.
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CHAPTER ELEVEN
0 < φ < π/2. The equations of motion are thus the same as for the inverted spherical
pendulum:
φ̈ − θ̇ 2 sin φ cos φ −
g
sin φ = 0
l
θ̈ + 2θ̇ φ̇ cot φ = 0.
Recall that Example 10.3 discussed the existence of a steady-state solution where the
pendulum hangs at a constant offset angle while circling around the vertical. Here
we show that for certain values of the angular momentum, an inverted steady-state
solution with a constant offset also exists.1
For the spinning case, we begin as for the spherical pendulum in Example 10.5.
The position and velocity of the rotor in spherical coordinates is
rP/O = ler
I
vP/O = l φ̇eφ + l θ̇ sin φeθ .
To find the angular momentum, we use the separation principle and write the total
angular momentum about O as the sum of the angular momentum of the rotor center
of mass P plus the angular momentum, her , about the rotor center of mass
hO = rP/O × mP IvP/O + her
I
= mP l 2φ̇eθ − mP l 2θ̇ sin φeφ + her .
(11.4)
We can now take the derivative of the total angular momentum as done in Example 10.5, using the angular velocity for the unit-vector derivatives, and set it equal
to the total moment about O, as in Eq. (11.2). The angular velocity of the spherical
frame, BS (Eq. (10.28)), is
I BS
ω
= −θ̇ sin φeφ + φ̇eθ + θ̇ cos φer ,
which we use to find the rate of change of the unit vectors in the angular momentum
equation in Eq. (11.4). The rotational kinematics are
Id dt
hO = mP l 2(φ̈ − θ̇ 2 sin φ cos φ)eθ − mP l 2 θ̈ sin φ + 2θ̇ φ̇ cos φ eφ
I
+ hθ̇ sin φeθ + hφ̇eφ .
As in Example 10.5, the only moment about O is due to gravity:
MO = rP/O × FP/O = ler × (−mP gez )
= −mP gler × (− sin φeφ + cos φer )
= mP gl sin φeθ .
1 Note
that the stability of these solutions is a separate question. Stability refers to the tendency of the
pendulum to return to the steady state if slightly perturbed away from it. Clearly the downward pendulum
equilibrium is stable but the inverted one is not. Examining stability is a more complicated question (though
the inverted gyropendulum is, in fact, stable); Chapter 12 presents an introduction to this subject.
MULTIPARTICLE AND RIGID-BODY DYNAMICS IN THREE DIMENSIONS
471
Equating the moment on the pendulum rotor about O with the rate of change of
angular momentum about O yields the following two scalar equations:
0 = 2mP l 2θ̇ φ̇ cos φ + mP l 2θ̈ sin φ + hφ̇
mP gl sin φ = mP l 2φ̈ − mP l 2θ̇ 2 sin φ cos φ + hθ̇ sin φ.
Solving for θ̈ and φ̈ produces the two equations of motion for the simplified gyropendulum:
hφ̇
=0
mP l 2 sin φ
(11.5)
g
hθ̇ sin φ
sin φ +
= 0.
l
mP l 2
(11.6)
θ̈ + 2θ̇ φ̇ cot φ +
φ̈ − θ̇ 2 sin φ cos φ −
We now look for a constant-offset solution. Is there a solution with constant
θ̇ = θ̇0 > 0 and constant φ = φ0, where 0 < φ0 < π/2? Setting φ̈ = φ̇ = 0 in Eq. (11.6)
yields
θ̇ 2 cos φ0 −
h
g
θ̇ + = 0.
l
mP l 2
(11.7)
Thus for θ̇ = θ̇0, the steady-state tilt of the pendulum is
g
h
−
.
φ0 = arccos
mP l 2θ̇0 l θ̇02
Such a solution exists as long as
0≤
g
h
−
≤ 1.
l θ̇02 mP l 2θ̇0
Alternatively, we can ask what the constant precession rate θ̇0 > 0 of the pendulum
is about the vertical for a given h and tilt angle φ0. This problem involves solving the
quadratic equation in Eq. (11.7) to obtain
4 cos φ0 g
h2
h
±
−
2
l
mP l
m2P l 4
θ̇0 =
.
2 cos φ0
Such a solution exists as long as
4 cos φ0g
h2
−
≥ 0.
l
m2P l 4
Thus we can state a condition on the angular momentum of the rotor for which an
inverted steady-state angle φ0 exists:
h≥
4 cos φ0gm2P l 3.
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CHAPTER ELEVEN
We conclude this section with a reminder regarding momentum conservation. Just
as in Chapters 7 and 9, Euler’s second law leads to the law of conservation of angular
momentum.
Law 11.1 Under the internal-moment assumption, the law of conservation of total
angular momentum states that, if the total external moment acting on a system
of particles about the origin O is zero, then the total inertial angular momentum
of the collection about O is a constant of the motion. Likewise, if the total
external moment about the center of mass G of the collection is zero, then the
inertial angular momentum of the collection about G is a constant of the motion.
11.2 Three-Dimensional Rotational Equations
of Motion of a Rigid Body
The primary objective in this section is to study the rotational motion of a rigid body
about its center of mass, employing the separation of angular momentum. Up to now
it may seem like there is not much new. The general expressions for the dynamics of
the angular momentum (Euler’s first and second laws) are exactly the same in three
dimensions. In fact, as described at the beginning of the chapter, in translation there is
nothing new. Newton’s second law (or Euler’s first law) is simply applied to the center
of mass in three dimensions, just as we did in two, and all results from Chapter 10
apply. The changes come when considering rotation.
Although the analysis of angular momentum in Chapters 7 and 9 used vectors
and thus was completely general, we only found the planar equations of motion.
This simplification is significant: by confining motion to a plane we constrained
the angular momentum to be always perpendicular to the plane. In other words, the
angular-momentum direction was fixed and, in fact, parallel to the angular velocity.
The number of rotational degrees of freedom collapsed from three to one and our
rotational equation of motion reduced to a single scalar equation. Here we allow the
particles or rigid body to move arbitrarily. The total angular momentum can thus
change both its direction and magnitude and is not necessarily parallel to the angular
velocity. Although the resulting equations are concise, these new degrees of freedom
result in amazingly complex and interesting behavior.
Our main observation is that Eq. (11.3), Euler’s second law, is not the most
convenient form of the vector equation of motion, as it differentiates the angular
momentum with respect to absolute space. Now that there are additional degrees of
freedom, the angular momentum can change in more complicated ways and, for most
problems, we are more interested in describing that change with respect to a body
frame. This is because (a) we are considering the motion of particles relative to a
moving body frame; (b) we are solving for the motion of a rigid body and it is easier
to express the angular momentum in its body frame; or (c) the applied moment is
most easily expressed in a body frame.
The first step, then, is to find a differential equation for the angular momentum with
respect to the body frame. To do so, we apply the transport equation to Eq. (11.3).
Because the angular momentum is a vector, we can use the transport equation and
MULTIPARTICLE AND RIGID-BODY DYNAMICS IN THREE DIMENSIONS
473
rewrite Eq. (11.3) to find the following alternative form of Euler’s second law in three
dimensions:
Bd
dt
hG + IωB × IhG = MG.
I
(11.8)
Eq. (11.8) is the key vector equation for all rotational rigid-body dynamics. It
is almost always the best place to start when solving three-dimensional dynamics
problems. Note that when we were confined to planar motion this extra step was
unnecessary, as the angular momentum was, by design, always aligned with the
angular velocity and thus the cross product between the two was zero. Now the use
of the transport equation has given us a much more useful form of Euler’s second law
for motion about the center of mass.
However, we can’t yet use Eq. (11.8) to solve problems exactly, as the threedimensional relationship between the angular momentum and the angular velocity
has not yet been discussed. We will do that soon, including how to incorporate the
Euler angles. Nevertheless, it is important not to lose sight, amidst all the mathematics,
that Eq. (11.8) is always the starting point. We are solving for the time evolution of
the angular momentum of a multiparticle system or a rigid body. The next subsection
discusses how this equation is approximated to qualitatively understand the motion
of a spinning body. The following two subsections modify Eq. (11.8) to provide exact
scalar equations of motion for the rotational trajectories in three dimensions.
11.2.1 Qualitative Analysis of a Spinning Body
Just as for the gyropendulum in Example 11.2, many physical systems involve a
rapidly spinning body or a system with a connected spinning rotor. It is often very
useful in these systems to develop a qualitative understanding of the expected behavior. By this we mean an approximate prediction of the dominant motion without an
exact solution of the detailed trajectories. For instance, in the gyropendulum, if the
rotor is spinning clockwise, is the pendulum going to precess to the right or to the
left? What direction does the earth’s spin axis move due to the torques exerted on it
by the sun and moon? If you hold a spinning bicycle wheel and tilt it to the left, do
you feel a clockwise or counterclockwise torque? Fortunately, all these questions can
be easily answered using Eq. (11.8) without resorting to finding and solving exact
equations of motion. The ability to qualitatively assess three-dimensional behavior is
an important engineering skill to develop.
The key point is that the angular momentum of a spinning body is dominated by the
contribution from the rapidly spinning part. All other motions make only a very small
contribution to the total angular momentum (or its change with time). Thus as a first
approximation, it is usually adequate to ignore all but the spin angular momentum. If
the spin rate is roughly constant, this observation suggests that we assume
Bd
dt
hG ≈ 0.
I
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CHAPTER ELEVEN
The result is that the equation of motion reduces to the simple vector equation
I B
ω × IhG ≈ MG.
(11.9)
Eq. (11.9) is not a differential equation! Our qualitative analysis says simply that
the angular velocity IωB of the body frame B is such that it satisfies the cross product
in Eq. (11.9). The only requirement is that we choose a body frame B where the spin
angular momentum is roughly constant and is large compared to the contributions
from the other motions. Getting comfortable with this sort of qualitative analysis of
problems is enormously helpful. It gives you a sense of what to expect when you
set out on the more complicated exact solution, which helps avoid mistakes. It also
provides an intuition that is indispensable when you move from analyzing a system
to synthesizing a design. Some examples will help clarify what we mean.
Example 11.3 Qualitative Analysis of the Gyropendulum
This example reconsiders the gyropendulum. In Example 11.2 we found the exact
equations of motion and then were able to work out several equilibrium solutions.
What if instead we had begun by just using Eq. (11.9) to determine the general
characteristics of the motion before solving for it exactly?
Figure 11.3a shows that the angular momentum is dominated by the spin of the
rotor:
I
hO ≈ her .
The moment on the gyropendulum about O is again, from the free-body diagram in
Figure 11.3c,
MO = mP gl sin φeθ .
Thus a good estimate of the rotational dynamics is found from Eq. (11.9). We have
I B
ω × her ≈ mP gl sin φeθ .
(11.10)
For this cross product to be satisfied, the angular velocity must be perpendicular
to eθ and have a component perpendicular to er . Thus there must be a component of
IωB that is in the −e direction (watch the sign!). This observation implies
φ
I B
ω ≈−
mP gl
m gl
sin φeφ = P (ez − cos φer ).
h
h
This qualitative analysis shows that, for a positive spin of the rotor, the body
frame fixed to the gyropendulum precesses about the ez axis in the counterclockwise
direction when viewed from above.
MULTIPARTICLE AND RIGID-BODY DYNAMICS IN THREE DIMENSIONS
475
Figure 11.4 Holding a spinning bicycle wheel on a swivel chair.
Example 11.4 The Spinning Bicycle Wheel
One of our favorite science-museum exhibits is the spinning bicycle wheel (Figure 11.4). In this exhibit, you sit in a swivel chair and pick up a large (usually somewhat
heavy) bicycle wheel. You then have a friend spin it up as fast as he or she can. Holding the wheel out in front of you, you tilt it either to the right or to the left. As you
do, you and the swivel chair will turn. It is an amazing yet simple demonstration of
gyroscopic motion. Using Eq. (11.9) we can quite easily determine the direction of
rotation of the chair.
We begin by connecting a body frame B = (G, b1, b2, b3) to the bicycle wheel,
with b1 along the axle and directed to your left, b2 pointed toward you, and b3 pointed
upward from your lap to complete the right-handed set. Note that we have not attached
the body frame to the spinning wheel, just to the axle. This practice is important and
is something done quite often in rigid-body analysis. For a symmetric body that is
spinning, we often define the body frame as an intermediate frame. That is because
we are not necessarily interested in the angle that the spinning rotor makes in absolute
space, but rather with the orientation of the rotor’s spin axis. This is perfectly fine (as
long as the rotor is symmetric). The angular velocity in Eq. (11.9) is that of the wheel
axle with respect to I .
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CHAPTER ELEVEN
Figure 11.5 Spinning top.
Once spun up, the bicycle wheel has a rather hefty angular momentum directed
in the b1 direction: IhG = hb1. When you tilt the bicycle wheel to the left (in the
direction of the angular momentum), you are applying a positive torque to it along
the b2 direction: MG = Mb2.2 An angular velocity of the body frame in absolute
space must be induced to satisfy Eq. (11.9). In other words, we have
I B
ω × hb1 = Mb2.
For this equation to be satisfied, the angular velocity must have a component IωB b3,
where IωB > 0. Thus there is a positive angular velocity pointed upward; the wheel
and the chair (with you in it) rotate counterclockwise when viewed from above. If
you turn the wheel to the right, which requires a negative moment, then the chair will
rotate clockwise.
Remember, this analysis is only approximate. A detailed treatment would use the
full vector equation of motion, employ multiple frames, and include reaction forces
in the chair. Nevertheless, if all we are interested in is predicting the direction the
chair rotates, the qualitative analysis works amazingly well!
Example 11.5 A Spinning Top
This example takes a qualitative look at the impact of friction on a top that spins
on a rounded (as opposed to pointed) tip (see Figure 11.5). Let C = (O, c1, c2, c3)
denote the body frame fixed to the top. Figure 11.6 illustrates frame C and the two
intermediate frames B and A that allow us to relate the orientation of C to an inertial
frame I using Euler angles ψ and θ. The third Euler angle φ describes the top’s
primary angle of rotation about b3 = c3; as for the bicycle wheel in the previous
example, we assume φ̇ 0. Thus we can approximate the top’s angular momentum
about its center of mass as IhG ≈ hGb3, where hG is a positive constant. Under this
approximation, the rate of change of IhG with respect to frame B is zero, and we can
gain insight into the behavior of the top using Eq. (11.9).
2 Note that this torque is an external moment on the wheel because, for you to be able to apply this moment
without counterrotating, you must push against the chair.
MULTIPARTICLE AND RIGID-BODY DYNAMICS IN THREE DIMENSIONS
I
θ
e3, a3
a2
C
b3, c3
c2
b3
a 2, b2
A
A
O
O
O
ψ
477
b1
a1
(a)
B
B
(b)
ϕ
c1
(c)
Figure 11.6 Reference frames for the spinning top. (a) Intermediate frame A. (b) Intermediate
frame B. (c) Body frame C.
To apply Eq. (11.9), we must first compute the (external) moment MG about the
center of mass. Since the weight of the top acts at its center of mass and therefore
does not exert a moment about it, the only forces that produce moments about G are
the contact forces from the ground. Let us assume that the top makes contact with the
ground at point O (as shown in Figure 11.7a) and d is the radius of curvature of the
bottom. In this case, the moment about G is
MG = rO /G × FO ,
where
rO /G = −lb3 + db1.
Using the free-body diagram in Figure 11.7b yields
FO = N a3 − μNb2,
A
B
a3
b3
G
G
θ
l
O
Na 3
O′
a1
b1
(a)
–mga3
–μNb2
O′
(b)
Figure 11.7 (a) Side view of spinning top in inertial frame A. (b) Free-body diagram used for
the spinning top.
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CHAPTER ELEVEN
where μ is the coefficient of friction. If, in addition, we assume that the bottom radius
d is significantly smaller than the height l of the center of mass relative to the bottom,
then the position of O relative to G is roughly rO /G ≈ −lb3. The transformation table
relating the unit vectors of frames A and B is
a1
a2
a3
b1
b2
cos θ
0
− sin θ
0
1
0
b3
sin θ
0
cos θ .
Therefore the moment about G is approximately
MG ≈ (−lb3) × (−N sin θb1 − μNb2 + N cos θ b3)
= lN sin θb2 − lμN b1.
(11.11)
Using IhG ≈ hGb3 and Eq. (11.11) in Eq. (11.9) to find IωB yields
I B
ω ≈−
lN sin θ
lμN
b1 −
b2 .
hG
hG
(11.12)
Now consider the angular velocity IωB in terms of the Euler angle rates:
I B
ω = ψ̇a3 + θ̇b2 = −ψ̇ sin θb1 + θ̇b2 + ψ̇ cos θb3.
(11.13)
Comparing Eqs. (11.12) and (11.13) reveals that ψ̇ ≈ lN/ hG > 0 and θ̇ ≈ −lμN/ hG
< 0. The first result implies that the top precesses counterclockwise at a constant rate
about the a3 axis. The second result implies that the angle θ that the top is tipped
away from vertical decreases at a rate proportional to the coefficient of friction, μ.
That is, friction causes the top to right itself! Although we are unable to predict the
b3 component of the angular velocity, we can predict the basic motion of the top.
11.2.2 The Moment of Inertia Tensor
Although a qualitative understanding of motion is tremendously useful, eventually
we do want to find scalar differential equations of motion that can be used to describe
rigid-body trajectories—that is, the evolution of its orientation—in terms of angularvelocity components and Euler angles. To do so, we follow a process similar to that
in Chapter 9 for planar motion: we find an expression for the angular momentum
in terms of the angular velocity. This then allows us to write the scalar component
magnitudes of Eq. (11.3) as differential equations in the coefficients of the angular
velocity expressed as components in the body frame (i.e., ω1, ω2, and ω3).
Chapter 9 used the planar assumption in several places. We now relax that assumption. The result is a bit more complicated, but it fully describes the three-dimensional
angular momentum of a rigid body. We begin, as before, with the equation
for the angular momentum of a rigid body (viewed as a collection of N particles)
about its center of mass G:
MULTIPARTICLE AND RIGID-BODY DYNAMICS IN THREE DIMENSIONS
I
hG =
N
479
mi ri/G × Ivi/G.
i=1
We use the transport equation on the inertial velocity of each particle, recognizing
that the body is rigid (so that there is no body-frame velocity) to find
I
vi/G = Bvi/G +IωB × ri/G.
=0
This result is substituted back into the formula for IhG to again find Eq. (9.15),
I
hG =
N
mi ri/G ×
ω × ri/G .
I B
i=1
We now deviate from our planar approach. Since motion is allowed in all three
directions, we can no longer simply substitute for the angular velocity and assume it
is fixed in the b3 direction. Instead, applying the vector triple cross-product rule (see
Appendix B) gives
I
hG =
N
mi (ri/G . ri/G)IωB − ri/G(ri/G . IωB ) .
(11.14)
i=1
Now things get a bit more sophisticated. How do we factor out the angular velocity
as was done in Chapter 9 for the planar case? We do it by introducing the tensor
product ⊗. The tensor product is described in Appendix B. Unlike the cross product,
which produces a vector described by three component magnitudes, the tensor product
of two vectors results in a second-rank tensor described by nine scalars. We use the
scalar tensor triple product described in Appendix B to rewrite the second term in
Eq. (11.14):
I
hG =
N
mi (ri/G . ri/G)IωB − (ri/G ⊗ ri/G) . IωB .
i=1
Don’t be intimidated by this new notation. It seems much more complicated than it
really is. Soon we will rewrite the results with matrices so that we can work with the
scalar component magnitudes directly. The tensor notation is convenient, however,
because it permits compact writing of the equations of motion in a vector form
independent of a specific reference frame, as has been done throughout the book.
We now factor out the angular velocity, as for the planar case, to obtain
N
I
hG =
mi (ri/G . ri/G)U − (ri/G ⊗ ri/G) .IωB ,
i=1
=IG
where U is the unity tensor and IG is the moment of inertia tensor. This equation is the
three-dimensional analog to Eq. (9.17), except here the moment of inertia is a tensor
rather than a scalar.
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CHAPTER ELEVEN
Definition 11.1 The moment of inertia tensor IG of a rigid body about its center
of mass G is the mass-weighted sum
IG =
N
mi (ri/G . ri/G)U − (ri/G ⊗ ri/G) .
i=1
For a continuous rigid body B,
IG =
B
(rdm/G . rdm/G)U − (rdm/G ⊗ rdm/G) dm.
Using Definition 11.1, the angular momentum of the body about its center of mass
can be written in the compact form
hG = IG . IωB .
I
(11.15)
Note, again, that the value of tensor notation is that it is frame independent. The
equation for the angular momentum in terms of the moment of inertia tensor in
Eq. (11.15) is the same no matter which frame we elect to express its components.
However, at some point we may need to choose a frame in which to express the vector
components. Appendix B describes how to find the components Iij of tensor IG in
frame B = (G, b1, b2, b3):
IG =
3
3 Iij bi ⊗ bj .
i=1 j =1
For instance, suppose that the position of particle k with respect to G has Cartesian
coordinates (xk , yk , zk )B . Then the discrete form in Definition 11.1 yields the following expressions for the components of the rigid body’s moment of inertia tensor in
that frame. When i = j ,
I11 =
N
mk yk2 + zk2
(11.16)
mk xk2 + zk2
(11.17)
mk xk2 + yk2 ,
(11.18)
k=1
I22 =
N
k=1
I33 =
N
k=1
MULTIPARTICLE AND RIGID-BODY DYNAMICS IN THREE DIMENSIONS
481
where we used rk/G = xk b1 + yk b2 + zk b3. The quantities I11, I22, and I33 are called
the moments of inertia. When i = j ,
I12 = I21 = −
N
mk xk yk
(11.19)
m k x k zk
(11.20)
m k yk z k .
(11.21)
k=1
I13 = I31 = −
N
k=1
I23 = I32 = −
N
k=1
These are called the products of inertia.
As an alternative to tensor notation, we now reprise this analysis using matrix
notation. In fact, it is in three-dimensional rigid-body analysis that matrix notation
comes in particularly handy. Recall that we can use matrix notation to write the
component magnitudes of a vector in a frame:
⎡
⎤
xk
[rk/G]B = ⎣ yk ⎦ ,
zk B
where, again, Cartesian coordinates locate particle k with respect to G in B. Appendix B similarly shows how to write the components of a tensor in matrix notation:
⎡
I11 I12
[IG]B = ⎣ I21 I22
I31 I32
⎤
I13
I23 ⎦ .
I33 B
This is known as the moment of inertia matrix. [IG]B is a 3 × 3 symmetric matrix
with elements calculated in a frame B fixed to the rigid body. The diagonal elements
are the moments of inertia of the rigid body (given by Eqs. (11.16)–(11.18)) and the
off-diagonal elements are the products of inertia (given by Eqs. (11.19)–(11.21)).
Appendix B shows how various vector and tensor operations can be written as
matrix products, allowing us to find expressions for the moment of inertia matrix in
terms of simple matrix operations. Thus we can use Definition 11.1 to write out the
discrete form of the moment of inertia tensor components in matrix notation:
[IG]B =
N
mi ([ri/G]TB [ri/G]B )I − [ri/G]B [ri/G]TB ,
i=1
where I is the 3 × 3 identity matrix.
For a continuous rigid body,
rdm/G2I − [rdm/G]B [rdm/G]TB dm.
[IG]B =
B
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CHAPTER ELEVEN
As with the scalar moment of inertia in Chapter 9, the moment of inertia matrix is
usually computed using a position-dependent density ρ(rdm/G) for the rigid body:
rdm/G2I − [rdm/G]B [rdm/G]TB ρ(rdm/G)dV,
[IG]B =
B
where dV is the differential volume element (see Appendix A).
As done earlier with the components of the tensor, let’s see what is going on by
writing out the matrix integral explicitly using the Cartesian coordinates (x, y, z)B
of mass element dm in frame B. Using these coordinates, the position [rdm/G]B is
[x y z]TB . Assuming a constant density ρ, the moment of inertia matrix is
⎡
y 2 + z2
⎣ −yx
[IG]B = ρ
B
−zx
−xy
x 2 + z2
−zy
⎤
−xz
−yz ⎦ dxdydz.
2
x + y2 B
(11.22)
The body-frame components of the angular momentum from Eq. (11.15) in matrix
notation are
[IhG]B = [IG]B [IωB ]B .
(11.23)
Example 11.6 Computing the Moment of Inertia of a Cube
This example uses the formulation of the moment of inertia matrix in Cartesian
coordinates to find the moment of inertia matrix of a constant-density cube about its
center. Consider a body frame B aligned with the edges of a cube and whose origin is
coincident with the center of the cube, as shown in Figure 11.8. Assume the length of
each side of the cube is 2l. Also assume that the mass of the cube is m, which implies
that ρ = m/(8l 3). Then the matrix in Eq. (11.22) becomes
m
[IG]B = 3
8l
l
−l
l
−l
⎡
y 2 + z2
⎣ −yx
−l
−zx
−xy
2
x + z2
−zy
l
⎤
−xz
−yz ⎦ dxdydz.
2
x + y2 B
B
G
2l
2l
2l
Figure 11.8 Computing the moment of inertia of a cube.
(11.24)
MULTIPARTICLE AND RIGID-BODY DYNAMICS IN THREE DIMENSIONS
483
B
G
R
Figure 11.9 Computing the moment of inertia of a sphere.
Eq. (11.24) is a 3 × 3 matrix of integrals. Let Iij denote the entry in the ith row and
j th column of this matrix, where i, j = 1, 2, 3. Evaluating the first integral yields
I11 =
m
8l 3
l
−l
l
l
2
(y 2 + z2)dxdydz = ml 2.
3
−l
−l
In fact, we have I22 = I33 = (2/3)ml 2 as well. Evaluating an off-diagonal integral
yields
I12 =
m
8l 3
l
−l
l
−l
l
−l
(−xy)dxdydz = 0.
In fact, all the off-diagonal integrals evaluate to zero. Therefore the moment of inertia
matrix relative to a frame perpendicular to the faces of a constant-density cube is
simply [IG]B = (2/3)ml 2I , where I is the 3 × 3 identity matrix.
Example 11.7 Computing the Moment of Inertia of a Sphere
The same procedure can be used to compute the moment of inertia matrix of a
constant-density sphere of radius R. We again introduce body frame B with origin at
the center of the sphere, as shown in Figure 11.9. Assuming again that the mass of
the sphere is m, then the density is ρ = 3m/(4π R 3). For this geometry, the moment
of inertia matrix is most easily computed using spherical coordinates. Using the
relationships from Eq. (10.4),
x = r cos θ sin φ
y = r sin θ sin φ
z = r cos φ,
we can replace the Cartesian coordinates in Eq. (11.22). The result is an integral
expression for the moment of inertia matrix:
R
π
3m
sin
φ
dφ
r 4dr
(11.25)
4πR 3 0
0
⎡
⎤
2
2
2
− 21 sin 2θ sin2 φ
− 21 cos θ sin 2φ
2π (sin θ sin φ + cos φ)
⎢
⎥
×
(cos2 θ sin2 φ + cos2 φ) − 21 sin θ sin 2φ ⎦ dθ,
− 21 sin 2θ sin2 φ
⎣
[IG]B =
0
− 21 cos θ sin 2φ
− 21 sin θ sin 2φ
sin2 φ
B
484
CHAPTER ELEVEN
where we used the volume element in spherical coordinates, dV = r 2 sin φdrdφdθ .
All products of inertia are zero because each includes an integral in θ over a full period
of sine or cosine. This is true no matter how we orient the frame chosen relative to
the sphere. Unlike a cube, where we can choose a specific geometry for the reference
frame B, for a sphere the moment of inertia matrix is always diagonal, no matter what
body frame is chosen. We’ll revisit this important observation later.
Completing the integrals over dr and dθ leaves (it helps to consult an integral
table)
⎤
⎡
0
0
(3 + cos 2φ)
3mR 2 π ⎣
⎦ sin φ dφ.
[IG]B =
0
0
(3 + cos 2φ)
40
0
2
0
0
4 sin φ B
Completing the remaining integrals over dφ results in the moment of inertia matrix
of a constant-density sphere: [IG]B = (2/5)mR 2I , where I is the 3 × 3 identity matrix.
The sphere is an example of an isoinertial body. An isoinertial body has identical
moment of inertia about any line through the center of mass and zero products
of inertia, no matter which body frame is chosen. It is an interesting fact that all
centrobaric3 bodies are also isoinertial. One of the consequences of a body being
isoinertial is that the angular momentum is always aligned with the angular velocity
in the body. This is not true in general for a three-dimensional rigid body, as we show
next.
Example 11.8 The Moment of Inertia and Angular Momentum
of a Rotating Rod
This example computes the moment of inertia tensor and angular momentum of the
long cylindrical rod shown in Figure 11.10. The rod has a circular cross section
of radius R and length l. We again introduce a body frame B with origin at the
center of mass G of the rod and with b3 aligned along the long axis of the rod.
Assuming uniform density and mass m, the density is ρ = m/(π R 2l). For this body,
cylindrical coordinates are best for computing the moments of inertia. Using our usual
relationship between cylindrical and Cartesian coordinates and dV = rdrdθdz, the
moment of inertia matrix is
R
2π
m
rdr
dθ
[IG]B =
πR 2l 0
0
⎡ 2 2
⎤
r sin θ + z2 −r 2 cos θ sin θ −rz cos θ
l/2
⎣ −r 2 cos θ sin θ r 2 cos2 θ + z2 −rz sin θ ⎦ dz.
×
−l/2
−rz cos θ
−rz sin θ
r2
B
Completing the integrals gives the moment of inertia matrix for the rod:
⎤
⎡ mR 2 ml 2
0
0
4 + 12
⎥
⎢
ml 2
mR 2
[IG]B = ⎣
0 ⎦.
0
4 + 12
0
3 Recall
0
mR 2
2
B
that for a centrobaric body the center of mass and the center of gravity coincide.
MULTIPARTICLE AND RIGID-BODY DYNAMICS IN THREE DIMENSIONS
485
B
–l
2
G
R
Figure 11.10 Computing the moment of inertia of a long cylindrical rod.
Unlike for the cube and sphere, the cylindrical rod has a different moment of inertia
about the z axis than it does about the x and y axes. Thus if the rod is rotating with
angular velocity [IωB ]B = [ω1 ω2 ω3]TB , the angular momentum of the rod is
⎤
⎡ mR 2 ml 2
( 4 + 12 )ω1
⎥
⎢ mR 2 ml 2
⎥
[IhG]B = ⎢
⎣ ( 4 + 12 )ω2 ⎦ ,
mR 2
B
2 ω3
which is not aligned with the angular velocity.
Often we will solve problems incorporating a “long thin rod”. This description
refers to a cylindrical rod with R l. In this case, we usually ignore the inertia about
the z-axis and approximate the moment of inertia matrix by
⎤
⎡ ml 2
0 0
12
⎥
⎢
ml 2
[IG]B = ⎣ 0
0⎦ .
12
0
0 0 B
11.2.3 The Rotational Equations of Motion
As for the planar case in Chapter 9, we can now take the expression for the angular
momentum of the rigid body in terms of the moment of inertia tensor and the angular
velocity in Eq. (11.15) and substitute into the body-frame vector equation of motion
in Eq. (11.8):
Bd I B
ω + IωB × IG . IωB = MG.
IG .IωB + IG .
dt
dt Bd =0
Note that we assumed the product rule applies to the scalar tensor triple product
and took the derivative in the frame of a tensor. This process highlights that we are
repeating the result found in Chapter 9, which is that the derivative of the moment of
inertia is zero, since it is defined in the body frame of the rigid body. That means the
rotational equation of motion simplifies to the important form:
IG .
Bd I B
dt
ω
+ IωB × IG . IωB = MG.
(11.26)
486
CHAPTER ELEVEN
Eq. (11.26) is the complete equation of motion for a rigid body rotating in absolute
space. Its components provide three differential equations for the angular speeds.
Note that it is a first-order differential equation in the angular-velocity component
magnitudes. Sometimes we may wish to substitute the kinematic equations of rotation
to obtain second-order differential equations in the Euler angles. Most of the time
we will leave them as first-order differential equations in the angular velocities and
combine them with the kinematic equations of rotation to get six first-order equations
that completely describe the rotational trajectory. Note that this is the same first-order
form used to numerically integrate equations of motion.
When converting Eq. (11.26) into scalar equations of motion, it is most convenient
to use matrix notation. We could just differentiate Eq. (11.23) directly and repeat our
derivation in matrix form. In fact, if you are finding tensor notation confusing, you
can always solve three-dimensional rigid-body rotation problems in matrix form; just
be sure you are careful to keep track of which frame of reference the components
are written in. We could alternatively use the matrix form of tensor operations in
Appendix B to directly convert Eq. (11.26). In either case, the matrix form of the
rotational equations of motion, expressed in a frame B fixed to the rigid body, is
[IG]B
Bd
dt
[IωB ]B + [IωB ×]B [IG]B [IωB ]B = [MG]B ,
(11.27)
where we have dropped the moment of inertia matrix derivative, since we assume the
moments and products of inertia are constant in the body frame. We have also used the
cross-product equivalent matrix to write the cross product with the angular velocity
in matrix notation as
⎤
⎡
0
−ω3 ω2
[IωB ×]B = ⎣ ω3
0
−ω1 ⎦ .
−ω2 ω1
0
B
We describe this convenient tool in Appendix B.
Example 11.9 A Spinning Symmetric Rigid Body
This example uses the matrix form of the rotational dynamics in Eq. (11.27) to derive
the exact equations of motion of the spinning rigid body shown in Figure 11.11a. The
body is symmetric and is rapidly spinning; it has a small moment of inertia about the
spin axis and equal moments of inertia about the other axes. We consider the case
where the body is subject to a constant external torque about its center of mass. Our
qualitative analysis (see Examples 11.3 and 11.4) using Eq. (11.9) predicts a steady
rotation of the spin-axis direction (check this!); here we verify this prediction using
a quantitative analysis. Observe that this is an excellent model of the earth’s geoid
experiencing small moments due to the gravitational attraction of the sun and moon.
We discuss this model a bit more at the end of the example.
Let C denote the body frame fixed to the rigid body with origin at its center of mass.
The intermediate frames A and B allow us to describe the orientation of C relative
to an inertial frame I using the Euler angles ψ and θ, as depicted in Figure 11.11b
MULTIPARTICLE AND RIGID-BODY DYNAMICS IN THREE DIMENSIONS
A
I
B
a3, e3
ψ
487
b3
b2
a2
C
Ω
θ
θ
G
G
ψ
(a)
G
b1, a1
a1
(b)
(c)
Figure 11.11 Spinning symmetric rigid body undergoing constant torque. (a) Coordinates.
(b) Frame A. (c) Frame B.
and c. The rigid body spins rapidly at angular rate about the b2 = c2 axis. (This
is a set of 3-1-2 Euler angles.) The angular velocity of C relative to B is therefore
BωC = b . The angular velocity of B relative to I is
2
I B
ω = ψ̇a3 + θ̇b1 = θ̇b1 + ψ̇ sin θb2 + ψ̇ cos θb3,
(11.28)
where we expressed IωB as components in frame B using the transformation table
between frames A and B:
a1
a2
a3
b1
b2
1
0
0
0
cos θ
sin θ
b3
0
− sin θ
cos θ .
The angular velocity of C relative to I is
I C
ω = θ̇b1 + (ψ̇ sin θ + )b2 + ψ̇ cos θb3.
We now compute the angular momentum IhG of the rigid body about its center
of mass and express it in the intermediate frame B fixed to the body symmetry axis.
(As in our qualitative analysis, choosing an intermediate frame to perform rotational
analysis of a spinning body often helps to simplify the calculations.) Because of the
physical symmetry, the body frame C yields a diagonal moment of inertia matrix,
⎡
I1 0
[IG]C = ⎣ 0 I2
0 0
⎤
0
0⎦ ,
I1 C
with identical first and third entries. Note that, because of the rotational symmetry
488
CHAPTER ELEVEN
about b2, the moment of inertia is unaffected by the spin, which implies that [IG]B =
[IG]C .4 In matrix notation, the angular momentum expressed in frame B is thus
⎤
⎡
I1θ̇
(11.29)
[IhG]B = [IG]B IωC = ⎣ I2(ψ̇ sin θ + ) ⎦ .
B
I1ψ̇ cos θ
B
We next assume a constant applied moment about the b1 axis of MG = −M0b1.
Depending on the specific application, this moment could arise from a number
of sources (e.g., aerodynamic loading on an airplane or gravitational torque on a
satellite). The important thing here is that it is a constant in the body frame.
Now we are ready to compute the rotational equations of motion using Eq. (11.27).
The only caveat is that we will produce the equations of motion using the transport
equation by considering the rate of change of the angular momentum with respect
to the intermediate frame B instead of the body frame C. (Either choice is valid; this
choice simplifies the calculation.) Also note that we will perform the matrix operations
with respect to frame B; the important thing is that we are consistent. (This is why we
use a frame subscript in our matrix notation!) The rotational dynamics with respect
to frame B are
Bd
dt
[IhG]B + [IωB ×]B [IhG]B = [MG]B .
Using Eqs. (11.28) and (11.29) yields the matrix equation
⎤
⎡
I1θ̈
⎣ I2 d (ψ̇ sin θ + ) ⎦
dt
I1(ψ̈ cos θ − ψ̇ θ̇ sin θ) B
⎤
⎤
⎡
⎡
⎤ ⎡
I1θ̇
−M0
0
−ψ̇ cos θ ψ̇ sin θ
+ ⎣ ψ̇ cos θ
0
−θ̇ ⎦ ⎣ I2(ψ̇ sin θ + ) ⎦ = ⎣ 0 ⎦ ,
0
I1ψ̇ cos θ
−ψ̇ sin θ
θ̇
0
B
B
B
which, in turn, yields the following rotational equations of motion:
I1θ̈ + (I1 − I2)ψ̇ 2 sin θ cos θ − I2ψ̇ cos θ = −M0
(11.30)
d ψ̇ sin θ + = 0
dt
(11.31)
I1ψ̈ cos θ + (I2 − 2I1)ψ̇ θ̇ sin θ + I2θ̇ = 0.
(11.32)
Eq. (11.31) is a perfect differential and thus integrates to yield a constant of the
motion,
ψ̇ sin θ + = C,
(11.33)
4 It is very common in the analysis of spinning symmetric rigid bodies to analyze only the orientation of
an intermediate body frame that is not spinning with the body, since we are typically not interested in the
spin orientation. As long as the body is symmetric with respect to this frame, so that the moments of inertia
are the same, this approach is valid.
MULTIPARTICLE AND RIGID-BODY DYNAMICS IN THREE DIMENSIONS
(a)
(b)
489
(c)
Figure 11.12 Motion of the rotating axis of a spinning body projected onto a sphere. Qualitative
analysis correctly predicts precession about the vertical axis, but fails to predict small
oscillations (nutation) that are evident in the exact solution. (a) ψ̇(0) = 0. (b) ψ̇(0) = −0.1rad/s.
(c) ψ̇(0) = 0.1 rad/s.
where C is a constant that depends on the initial conditions. In fact, Eq. (11.33) implies
that the component of the angular momentum about the spin axis is conserved. We
use this observation to simplify Eqs. (11.30) and (11.32):
I1θ̈ + I1ψ̇ 2 sin θ cos θ − I2C ψ̇ cos θ = −M0
(11.34)
I1ψ̈ cos θ + I2C θ̇ − 2I1ψ̇ θ̇ sin θ = 0.
(11.35)
Eqs. (11.34) and (11.35) are two differential equations of motion that can be
integrated to yield trajectories for ψ and θ. The resulting solution exhibits precession
(the slow rotation of the symmetry axis in ψ) and nutation (the wobble of the
symmetry axis in θ). Figure 11.12 illustrates how precession and nutation depend
on the initial value of ψ̇ by plotting the projection of the tip of the symmetry axis
b2 onto a sphere. The plots were generated using matlab to integrate Eqs. (11.34)
and (11.35) with (0) = 20 rad/s, θ(0) = 30◦, I1 = 10 kg-m2, I2 = 1 kg-m2, and
M0 = 5 N-m. Note how the shape of the projected trajectory depends on ψ̇(0).
Precession and nutation are exhibited by any spinning body undergoing a constant
torque. For example, as we alluded to at the beginning of the example, the earth
can be modeled as a spinning rigid body that is being torqued by the sun and moon
(due to slight asphericities in its shape). Its symmetry axis precesses with a period of
approximately 26,000 years, known as the precession of the equinoxes. There is also a
small nutation about the precession of roughly 9 arcseconds, known as the Chandler
wobble.
Although the exact solution to these equations of motion predicts behavior that is
significantly more complicated than our qualitative analysis, the qualitative solution
was approximately correct. Recall that we predicted the rigid body would precess at a
roughly constant rate. Our simulations support this prediction. In fact, we can find an
approximate solution for precession and nutation by considering small deviations of
the spin axis. Assuming that θ and θ̇ are both small, the equations of motion reduce to
I1θ̈ + I1ψ̇ 2θ − I2C ψ̇ = −M0
I1ψ̈ + I2C θ̇ = 0.
(11.36)
(11.37)
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Eq. (11.37) is again a perfect differential that can be integrated to yield
ψ̇ = −
I2
Cθ + k,
I1
where k is a constant that depends on the initial conditions. Substituting into
Eq. (11.36) yields
θ̈ + k +
2
I22
I12
C
2
θ =−
M 0 I2
+ Ck,
I1
I1
which is the equation for simple harmonic motion! Thus θ oscillates (nutates) about
a constant offset, while ψ precesses at a roughly constant rate with small oscillations.
Eq. (11.33) shows that the spin rate will oscillate by a small amount about the nominal
value as well.
11.2.4 Principal Axes and Euler’s Equations
Although the scalar equations represented by the matrix form of the equation of
motion in Eq. (11.27) work for every problem, in their most general form (that is, for
an arbitrary body frame and moment of inertia matrix), they can be quite complicated.
The moment of inertia matrix has nine elements. Because it is a symmetric matrix,
only six elements are independent. Nonetheless, the various matrix multiplications
can result in rather cumbersome equations. Fortunately, there is an important and
common simplification that makes the resulting equations easier to write down and
easier to solve: we can assume that the moment of inertia matrix is diagonal. In fact,
we did that in the previous example. Finding a general rule involves the concept of
principal axes.
For any rigid body, we can always find a body frame Bp within which the products
of inertia are zero and the moment of inertia matrix is diagonal. These special axes are
called principal axes. Many bodies have obvious symmetries, such as in Example 11.9;
in these cases the principal axes correspond to the symmetry axes of the body. For
situations where it is easier to calculate the moments and products of inertia in other
axes, it is always possible to transform to principal axes through an ordered rotation.
We discuss how to do that in Section 11.5. As shown there, the rotation matrix
consists of the eigenvectors of the moment of inertia matrix and the principal moments
of inertia are the eigenvalues. For now, it is important just to remember that, for
most problems, it is wise to choose a principal-axes body frame. This frame almost
always corresponds to symmetry axes of the body. Appendix D tabulates the principal
moments of inertia and principal axes for a number of simple rigid bodies.
Choosing a principal-axes body frame greatly simplifies the equations of motion
for the rigid body. Since the moment of inertia matrix takes the simple diagonal form
⎡
I1 0
⎣
[IG]Bp = 0 I2
0 0
⎤
0
0 ⎦,
I3 Bp
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491
substituting into the matrix form of the rotational equation of motion gives
⎤ ⎡
⎤
⎡
ω̇1
I1 0 0
⎣ 0 I2 0 ⎦ ⎣ ω̇2 ⎦
0 0 I3 Bp ω̇3 Bp
⎤ ⎡
⎡
⎤ ⎡
⎤
⎡
⎤
I1 0 0
ω1
0
−ω3 ω2
M1
+ ⎣ ω3
0
−ω1 ⎦ ⎣ 0 I2 0 ⎦ ⎣ ω2 ⎦ = ⎣ M2 ⎦ , (11.38)
0 0 I3 Bp ω3 Bp
−ω2 ω1
0
M3 Bp
Bp
where Mi bi , i = 1, 2, 3 are the components of MG in Bp .
The matrix equation in Eq. (11.38) is equivalent to the following three scalar
equations for the angular velocity components in Bp :
I1ω̇1 + (I3 − I2)ω2ω3 = M1
I2ω̇2 + (I1 − I3)ω1ω3 = M2
I3ω̇3 + (I2 − I1)ω1ω2 = M3.
(11.39)
These three scalar equations are called Euler’s equations. They are the cornerstone of
most rigid-body rotation problems. There is no new physics here—Euler’s equations
are just another way to write the angular-momentum equation from Eq. (11.8) in
terms of component magnitudes. Collectively these equations are the scalar version
of Euler’s second law.
Example 11.10 Torque-Free Motion
One of the most important examples of rigid-body motion and the use of Euler’s
equations is the study of torque-free motion of a spinning rigid body—the rotational
motion of a rigid body in free space in the absence of any torques on it. For the planar
situations studied in Chapter 9, this example is completely trivial; the rigid body either
will not rotate or will spin at a constant angular velocity about the fixed axis. However,
in three dimensions, the motion results in some surprising effects. Even though the
angular momentum must be fixed in magnitude and direction in absolute space (by
conservation of angular momentum), the angular velocity (and the body frame) can
change orientation with respect to the angular momentum.Note that this problem well
approximates a wide variety of physical systems, such as rotating satellites, falling
objects, and planetary rotation.
We begin by writing Euler’s equations (Eq. (11.39)) in the absence of applied
moments or torques:
I1ω̇1 + (I3 − I2)ω2ω3 = 0
I2ω̇2 + (I1 − I3)ω1ω3 = 0
I3ω̇3 + (I2 − I1)ω1ω2 = 0.
Without loss of generality, we consider a rigid body spinning about its b3 axis. Thus
we label the angular velocity about the 3-axis as ωs + ω3, where ωs is a (large) constant
spin rate and ω3 is a small perturbation. We also assume that the other two angular
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velocity component magnitudes, ω1 and ω2, are small compared to ωs . Dropping the
small second-order products ω1ω2, ω1ω3, and ω2ω3, Euler’s equations become
I1ω̇1 + (I3 − I2)ωs ω2 ≈ 0
(11.40)
I2ω̇2 + (I1 − I3)ωs ω1 ≈ 0
(11.41)
I3ω̇3 ≈ 0.
(11.42)
These equations (with ωs = ω3) are exact for a symmetric rigid body (where I2 = I1).
That is, for a symmetric torque-free rigid body, the motion we find in this example is
true for all values of ω1 and ω2, not just for small deviations.
Eq. (11.42) shows that the spin of the body about the 3-axis stays roughly constant
(and exactly so for a symmetric body). For the angular rates about the other two axes,
we differentiate Eqs. (11.40) and (11.41) and combine the result to find the separate
equations of motion,
ω̈1 + ωn2 ω1 = 0
ω̈2 + ωn2 ω2 = 0,
where
ωn2 =
(I3 − I2)(I3 − I1)ωs2
I1I2
(11.43)
.
As long as ωn2 = 0, these two equations of motion represent simple harmonic
motion in ω1 and ω2! Thus the angular rates about the 1- and 2-axes remain small
and oscillate about their small initial values at the same frequency ωn. The spin axis,
rather than being directly along the body 3-axis, is slightly offset from the 3-axis and
rotates about it. From Eq. (11.43) we see that this occurs when I3 > I1 and I3 > I2,
or I3 < I1 and I3 < I2. In other words, when the body spins about either the principal
axis of maximum moment of inertia (the major axis) or the principal axis of minimum
moment of inertia (the minor axis) its motion is stable (Figure 11.13). This motion of
the angular velocity about the 3-axis of the body (the symmetry axis for symmetric
rigid bodies) is often called free-body precession.
ωB
I
b2
I
ωB
B
b1
B
I
ωB
b3
(a)
b1
b3
B
b1
b3
b2
b2
(b)
(c)
Figure 11.13 Torque-free motion. (a) Spin about major axis. (b) Spin about minor axis. (c) Spin
about intermediate axis.
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b1
493
b2
B
b3
ex
O
ey
I
ez
Figure 11.14 Inertial frame and body frame of an airplane.
When the spin occurs about the intermediate axis of inertia, the equations of motion
become unstable and produce exponential growth of the angular-velocity components
over time. The motion rapidly deviates from a simple spin about the 2-axis. You should
remember this fundamental result of rigid-body analysis: torque-free spin about the
intermediate axis of inertia of a rigid body is unstable.
Example 11.11 The Rotational Equations of Motion of an Airplane
The previous chapter (Examples 10.4, 10.10, and 10.14) solved for the translational
and rotational kinematics of an airplane in flight but only for the translational equations of motion. We complete the treatment here with the rotational equations of
motion. This turns out to be straightforward, using the same matrix treatment that led
us to Euler’s equations. Unfortunately, we can’t simply substitute into Euler’s equations because the airplane’s body frame B, shown in Figure 11.14, does not coincide
with principal axes. Rather, the moment of inertia matrix in the body frame is
⎡
0
I1
[IG]B = ⎣ 0
I2
−I13 0
⎤
−I13
0 ⎦.
I3
B
(11.44)
Most aircraft are symmetric about the vertical plane (containing b1 and b3), thus
making I12 = I23 = 0. However, they don’t have front-to-back symmetry, which
implies nonzero products of inertia. Of course, it is certainly possible to find the
principal axes for any aircraft, but they would differ for each plane and would not
coincide with the obviously convenient geometric directions of the conventional body
frame. Because the body frame in Figure 11.14 is so convenient for expressing the
forces and moments and describing the configuration of the plane, it is universally
used at the expense of having slightly more complicated rotational equations of
motion.
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CHAPTER ELEVEN
Using the conventional notation for the angular-velocity components, [IωB ]B =
[p q r]TB (see Example 10.9), the equations of motion in matrix form for the symmetric airplane are
⎤ ⎡ ⎤
0 −I13
ṗ
I1
⎣ 0
I2
0 ⎦ ⎣ q̇ ⎦
−I13 0
I3
ṙ B
B
⎡
⎤ ⎡
0
I1
0 −r
q
⎣
⎣
⎦
+
0
I2
r
0 −p
−q p
0 B −I13 0
⎡
⎤
⎤ ⎡ ⎤
⎡
−I13
p
M1
0 ⎦ ⎣ q ⎦ = ⎣ M2 ⎦ ,
M3 B
I3
r B
B
where Mi is the ith component of the applied moment in the body frame. Solving for
ṗ, q̇, and ṙ requires multiplying both sides from the left by the inverse of the moment
of inertia matrix. We compute
⎡
⎢
⎢
[IG]−1
=
⎢
B
⎣
1
2 /I
I1−I13
3
0
I13
2
I1I3−I13
0
1
I2
0
I13
2
I1I3−I13
0
1
⎤
⎥
⎥
⎥.
⎦
2 /I
I3−I13
1
B
The rotational equations of motion for the symmetric airplane are
ṗ =
1
2
I
(I
−
I
)qr
−
I
I
pq
+
I
(I
−
I
)pq
+
I
qr
3 3
2
3 13
13 2
1
13
2
I1I3 − I13
1
I3M1 + I13M3
2
I1I3 − I13
M
1 q̇ =
(I1 − I3)pr + I13(p 2 − r 2) + 2
I2
I2
1
2
I
ṙ =
(I
−
I
)pq
+
I
I
qr
+
I
(I
−
I
)qr
−
I
pq
1 2
1
1 13
13 3
2
13
2
I1I3 − I13
+
+
1
I13M1 + I1M3 .
2
I1I3 − I13
These three equations are combined with the inverse kinematic equations from
Example 10.11 (Eq. (10.50)) to obtain the set of six first-order differential equations
that describe airplane orientation. The inverse kinematic equations are
ψ̇ = (q sin φ + r cos φ) csc θ
(11.45)
θ̇ = q cos φ − r sin φ
(11.46)
φ̇ = (q sin φ + r cos φ) tan θ + p.
(11.47)
Note that these six equations differ from most equations of motion we have found
in that they do not consist of second-order differential equations in the coordinates
MULTIPARTICLE AND RIGID-BODY DYNAMICS IN THREE DIMENSIONS
495
((ψ, θ, φ)IB in this case). Rather than use the coordinates and their rates to describe
the rotational state, we use the coordinates and the angular-velocity components.
Nevertheless, these six first-order differential equations can be integrated using the
same numerical approach as we have used throughout the book.
For completeness, we repeat the six translational equations (kinematic and kinetic)
from Example 10.14 (Eqs. (10.57)–(10.59)):
u̇ = rv − qw + F1/mG
v̇ = pw − ru + F2/mG
ẇ = qu − pv + F3/mG
and
⎡ ⎤
⎤
u
ẋ
⎣ ẏ ⎦ = IC B ⎣ v ⎦ .
w B
ż I
⎡
These 12 differential equations completely describe the six degrees of freedom of
an airplane in flight. Of course, we can’t yet integrate them (even numerically), as we
have not yet found the forces and moments on the airplane. We discuss these further
in Chapter 12.
11.3 The Moment Transport Theorem and the Parallel
Axis Theorem in Three Dimensions
We now reexamine the motion of a rigid body relative to an arbitrary point Q other
than the center of mass G, as done for the planar case in Chapter 9. This means
revisiting the discussions in Sections 9.4.1 and 9.4.2, where we derived the moment
transport theorem and the parallel axis theorem. The treatment in Section 9.4.1 was
again completely general in terms of vectors in three dimensions. Thus our final result
for the moment transport theorem in Eq. (9.32) holds without change:
MQ = MG − rQ/G ×
N
Fi(ext).
(11.48)
i=1
In words, given the moment on a rigid body about its center of mass, Eq. (11.48) is
used to find the resulting moment about some other point Q of the rigid body.
For the parallel axis theorem we are not so fortunate; treating that in three dimensions is more involved and requires a bit of care. In particular, we need to go back to
the beginning of the derivation in Section 9.4 and recompute the angular momentum.
Nevertheless, we once again take advantage of the fact that the treatment in Chapter 7
was entirely general. Thus, just as in Chapter 9 for planar rigid bodies, the equation of
motion for the angular momentum of the rigid body about Q is given by Eq. (7.17):
Id
dt
hQ = MQ + rQ/G × mGIaQ/O ,
I
(11.49)
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CHAPTER ELEVEN
where we omitted the superscript (ext), as we are treating a rigid body. As before,
Eq. (11.49) is most useful when the rigid body is pinned at Q and the resulting equation
reduces to Euler’s second law for the total angular momentum. Also, just as for the
angular momentum about the center of mass, we can rewrite this equation of motion
using the transport equation to be in terms of the body-frame derivative:
Bd
dt
hQ + IωB × IhQ − rQ/G × mGIaQ/O = MQ.
I
(11.50)
The final step is to find an expression for the angular momentum IhQ, so we
can develop a modified form of Euler’s equations. What we would like to find is
a relationship between IhQ and the angular velocity, somehow incorporating the
moment of inertia matrix. When treating motion in two dimensions in Section 9.4,
this relationship turned out to be relatively straightforward. We could write down IhQ
in terms of a scalar moment of inertia about the point Q and relate this moment of
inertia to the one about the center of mass G using the parallel axis theorem. Our
objective here is to generalize this to three dimensions.
The most obvious approach is to start with the definition of IhQ and manipulate it
in the same way done for IhG earlier. Thus, as in Section 9.4, we start with
I
hQ =
N
ri/Q × mi Ivi/Q
i=1
and go through the same series of substitutions (using the transport equation) and
manipulations as done in the last section. That is, we use the transport equation to
substitute IωB × ri/Q for the inertial velocity and use the triple vector cross product
rule to obtain the analog of Eq. (11.14). Using the scalar tensor triple product to find
the angular momentum in terms of a moment of inertia about Q results in
I
hQ = IQ . IωB ,
(11.51)
where the moment of inertia tensor about Q is similar to that about G:
IQ =
N
mi (ri/Q . ri/Q))U − (ri/Q ⊗ ri/Q) .
(11.52)
i=1
For a continuous body,
IQ =
B
(ri/Q . ri/Q)U − (ri/Q ⊗ ri/Q) dm.
Just as for the center-of-mass version, we substitute Eq. (11.51) into Eq. (11.50) for
the equations of motion of the rigid body, but with the added term due to the inertial
motion of Q. In tensor form, we have
IQ .
Bd I B
dt
ω
+ IωB × IQ . IωB − rQ/G × mGIaQ/O = MQ.
(11.53)
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497
In matrix form, Eq. (11.53) becomes
[IQ]B
Bd
dt
[IωB ]B + [IωB ×]B [IQ]B [IωB ]B
− mG[rQ/G×]B [IaQ/O ]B = [MG]B .
(11.54)
As in the planar case, if the body is pinned in absolute space at Q, then the
correction term is zero and we have the same equations of motion as before, except that
the moment of inertia matrix is taken relative to Q (which is now inertially fixed) rather
than G. If the inertia matrix is also diagonal, then we again have Euler’s equations. It
is important to remember that, in this derivation, we assumed Q is fixed to the body,
B
which implies dtd (ri/Q) = 0 for all points i on the rigid body. This assumption implies
the moment of inertia matrix about Q is fixed in B and thus factors out of the time
derivative.
To find an expression for IQ, we begin once again by recognizing that the developments in Chapter 7 were independent of dimension. Our vector equations apply
equally well in three dimensions. Thus we can use the same equation as in Section 9.4.2 when discussing the (planar) parallel axis theorem; that is, the expression
for IhQ in terms of IhG is
I
hQ = IhG + mGrQ/G × IvQ/G.
(11.55)
We replace the inertial velocity of point Q by means of the transport equation,
recognizing that the body-frame velocity of Q is zero (because Q is attached to the
body) to obtain
hQ = IhG + mGrQ/G ×
I
ω × rQ/G .
I B
We again use the triple vector cross product rule to transform this expression:
hQ = IhG + mG (rQ/G . rQ/G)IωB − (rQ/G . IωB )rQ/G .
I
This equation looks familiar. It is very similar to the expression in the summation
over mass particles in our original derivation of the moment of inertia tensor (prior
to Definition 11.1). We can thus use the same tensor product identity on the second
term (see Appendix B) to rewrite it as
hQ = IG . IωB + mG (rQ/G . rQ/G)IωB − (rQ/G ⊗ rQ/G) . IωB ,
I
where we have substituted from Eq. (11.15) for the angular momentum about the
center of mass.
Finally, we can factor out the angular velocity to get our final expression for the
angular momentum about Q,
hQ = IG + mG (rQ/G . rQ/G)U − (rQ/G ⊗ rQ/G) . IωB ,
I
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CHAPTER ELEVEN
where U is the unity tensor. Comparing this result to Eq. (11.51) for the angular
momentum about Q shows that the moment of inertia tensor about Q is
IQ = IG + mG (rQ/G . rQ/G)U − (rQ/G ⊗ rQ/G) .
(11.56)
Eq. (11.56) is the three-dimensional version of the parallel axis theorem. It provides
a formula for computing the moment of inertia tensor about an arbitrary point Q fixed
to a rigid body, given the moment of inertia tensor about the center of mass G and the
position of Q with respect to G. It is used the same way as the planar form was used
in Section 9.4.2.
We can also write the parallel axis theorem in matrix form, which is useful when
actually doing computations:
[IQ]B = [IG]B + mG rQ/G2I − [rQ/G]B [rQ/G]TB ,
(11.57)
where I is the 3 × 3 identity matrix.
Example 11.12 Transporting the Space Shuttle atop a 747
The space shuttle was frequently transported atop a specially equipped Boeing 747
aircraft as shown in Figure 11.15. To readjust the airplane’s control systems, the
total moment of inertia of the combined system is needed. Rather than compute that
quantity directly, it is easier to use the known inertia tensors of the two vehicles.
Each has a moment of inertia matrix relative to their individual centers of mass, as
in Eq. (11.44). Let [IGp ]Bp be the moment of inertia matrix of the plane in frame Bp
] be the moment of inertia matrix of the shuttle
relative to its center of mass, and [IG
s Bs
in frame Bs relative to its center of mass. We have
⎡
0
I1
[IGp ]Bp = ⎣ 0
I2
−I13 0
⎤
−I13
0 ⎦
I3
Bp
⎡
and [IG
]Bs
s
0
I1
=⎣ 0
I2
−I13
0
⎤
−I13
0 ⎦ .
I3
Bs
Because of the symmetries of both the plane and the shuttle, we assume that the center
of mass of the shuttle is offset from that of the aircraft along the plane of symmetry
and that the two body frames are not rotated with respect to each other. What we
seek is the composite moment of inertia matrix expressed in frame Bp relative to the
center of mass of the aircraft. To find it, we first need to find the shuttle’s moment
of inertia matrix relative to the plane’s center of mass and then add it to the plane’s
moment of inertia matrix.
We begin by writing the position of the plane’s center of mass relative to the
shuttle’s:
⎡ ⎤
d1
⎣
[rGp/Gs ]Bs = 0 ⎦ .
d3 Bs
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499
Figure 11.15 Space shuttle being transported on a Boeing 747. Image courtesy of Shutterstock.
We then find the moment of inertia of the shuttle relative to Gp using Eq. (11.57):
⎡
0
I1
=⎣ 0
I2
−I13
0
[IG
]Bs
p
⎡
⎤
−I13
0
d32
d12 + d32
0 ⎦ + ms ⎣ 0
−d1d3
I3
0
Bs
⎤
−d1d3
0 ⎦ .
d12
Bs
Since the two body frames align, the shuttle’s moment of inertia matrix is the same
expressed in Bp . Adding it to the plane’s moment of inertia matrix gives the total
moment of inertia matrix of the plane plus shuttle relative to the center of mass of the
plane:
⎡
⎢
(tot)
[IG
]Bp = ⎣
p
I1 + I1 + ms d32
0
−m d d
−I13 − I13
s 1 3
0
I2 + I2 + ms (d12 + d32 )
0
−m d d
−I13 − I13
s 1 3
0
I3 + I3 + ms d12
⎤
⎥
⎦.
Example 11.13 Exact Solution of a Demonstration Gyroscope
This example uses the matrix form of the rotational dynamics in Eq. (11.54) and
the parallel axis theorem in Eq. (11.57) to derive the exact equations of motion of
a demonstration gyroscope depicted in Figure 11.16a. The gyro consists of a large
spinning rotor at one end of a (massless) rod, able to spin on its axis, and a small
I ψ
g
B
G
θ
Q
m
I
P
Ω
M
–(m + M)ga3
(a)
B
a3, e3
b3
b2
a2
θ
Q
ψ
Q
a1
A
(b)
A
b1, a1
(c)
Figure 11.16 Exact solution of a demonstration gyroscope. (a) Coordinates. (b) Frame A.
(c) Frame B.
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CHAPTER ELEVEN
mass whose position is adjustable at the other end of the rod. We treat the entire
arrangement as a single rigid body. When the small mass is placed so that the support
is exactly at the center of mass G, there are no moments or torques on the system
and it balances. However, when the small mass is moved to misalign the center of
mass, gravity produces a moment about the pinned point Q, resulting in interesting
and complex motions of the gyro. Using a qualitative analysis as in Section 11.2.1,
we predict a steady counterclockwise rotation of the gyro axis (check this!). In what
follows, we perform a more careful quantitative treatment to verify this prediction
and to study other aspects of the motion.
Figure 11.16a defines two frames: an inertial frame I located at the top of the
support post at point Q and an intermediate body frame B fixed to the gyro shaft
but not spinning with the wheel. We are interested in the orientation of B in I and
the resulting trajectory. Since we are not concerned with the rotation of the rotor,
there are only two degrees of freedom in this problem and the orientation of B in I
is described by the two Euler angles, ψ and θ, shown in Figure 11.16a. These are
the first two angles in a 3-1-3 Euler rotation set. The intermediate frame A shown in
Figure 11.16b allows us to find the angular velocity of B relative to I ,
I B
ω = ψ̇a3 + θ̇b1 = θ̇b1 + ψ̇ sin θb2 + ψ̇ cos θb3,
(11.58)
where we have expressed IωB in terms of components in frame B using the transformation table between frames A and B:
a1
a2
a3
b1
b2
1
0
0
0
cos θ
sin θ
b3
0
− sin θ
cos θ .
To find the angular momentum of the gyro relative to Q, we use the same approach
as in Examples 11.4 and 11.5 and add the spin angular momentum of the rotor, hb2,
to the angular momentum due to the motion of the rigid body about Q:
[IhQ]B = [IQ]B [ IωB ]B + [hb2]B ,
where [IQ]B is the moment of inertia matrix relative to Q given by the parallel axis
theorem in Eq. (11.57). As in Example 11.9, we are ignoring the spin angle of the rotor
and only accounting for its angular-momentum contribution. We also are solving for
motion relative to a point that is not the center of mass, which necessitates the use of
the parallel axis theorem.
Again, using the same approach followed in Example 11.9, we are only interested
in the orientation of the intermediate body frame. We showed there that this approach
was valid as long as the body was symmetric about the spin axis. That is the case here
too, which allows us to write the moment of inertia matrix of the rotor about its center
of mass P as the following diagonal matrix in frame B:
⎡
⎤
I1 0 0
[IP ]B = ⎣ 0 I2 0 ⎦ ,
0 0 I1 B
MULTIPARTICLE AND RIGID-BODY DYNAMICS IN THREE DIMENSIONS
501
which has identical first and third entries because of the symmetry of the rotor. We
now use the parallel axis theorem in Eq. (11.57) to find the moment of inertia of the
rotor about point Q. Adding the contribution to the rigid-body inertia of the small
mass, we obtain
⎡
I1 0
[IQ]B = ⎣ 0 I2
0 0
⎤
⎡ 2
0
l
0 ⎦ +M ⎣ 0
I1 B
0
0
0
0
⎤
⎡ 2
0
l
0 ⎦ +m⎣ 0
l2 B
0
0
0
0
⎤
0
0 ⎦,
l 2 B
(11.59)
where l = rP/Q is the distance of the rotor from the attachment point and l = rm/Q
is the distance of the small mass from the attachment point.
In matrix notation, the angular momentum expressed in frame B is
⎤
I1θ̇
[IhQ]B = [IQ]B [ IωB ]B + [hb2]B = ⎣ I2ψ̇ sin θ + h ⎦ ,
I1ψ̇ cos θ
B
⎡
(11.60)
where I1 = I1 + Ml 2 + ml 2.
The next step is to compute the moment about Q. Here we refer to Examples 9.4
and 9.12, which showed that the moment about the center of mass of a rigid body due
to a uniform gravity field is zero and thus that the moment on a compound pendulum
is just the moment about Q due to the total force at the center of mass acting about
Q. In those examples we never used the planar restriction, so they apply unchanged
here. The moment is thus easily computed using the moment transport theorem in
Eq. (11.48), where we ignore the force at Q because it cancels:
MQ = rG/Q × FG = lGb2 × (−mGg sin θb2 − mGg cos θb3)
= −mGglG cos θb1,
(11.61)
where mG = m + M and lG = rG/Q.
Finally, we find the rotational equations of motion by using Eq. (11.50) with
Ia
Q/O = 0 and expressed in matrix notation:
Bd
dt
[IhQ]B + [IωB ×]B [IhQ]B = [MQ]B .
Using Eqs. (11.58)–(11.61) yields the matrix equation
⎤
I1θ̈
⎣ I2 d (I2ψ̇ sin θ + h) ⎦
dt
I1(ψ̈ cos θ − ψ̇ θ̇ sin θ ) B
⎤
⎤
⎡
⎡
⎤ ⎡
I1θ̇
−mGglG cos θ
0
−ψ̇ cos θ ψ̇ sin θ
⎦,
+ ⎣ ψ̇ cos θ
0
0
−θ̇ ⎦ ⎣ I2ψ̇ sin θ + h ⎦ = ⎣
I1ψ̇ cos θ
0
−ψ̇ sin θ
θ̇
0
B
B
B
⎡
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CHAPTER ELEVEN
which, in turn, yields the following rotational equations of motion:
I1θ̈ + (I1 − I2)ψ̇ 2 sin θ cos θ − hψ̇ cos θ = −mGglG cos θ
(11.62)
d
(I2ψ̇ sin θ + h) = 0
dt
(11.63)
I1ψ̈ cos θ + (I2 − 2I1)ψ̇ θ̇ sin θ + hθ̇ = 0.
(11.64)
Observe the similarity between this result and the one we obtained in Example 11.9
(Eqs. (11.30)–(11.32)). In particular, Eq. (11.63) is a perfect differential, indicating
that I2ψ̇ sin θ + h is a constant of the motion. The demo gyro exhibits the same kind
of precession and nutation characteristics that are illustrated in Figure 11.12. We defer
further analysis of this observation to Tutorial 11.2.
To conclude, we note that the parallel axis theorem in Eq. (11.56) is only valid for
transporting the moment of inertia tensor from about the center of mass to about an
arbitrary point fixed in the body. It cannot be used to transport the moment of inertia
from one arbitrary point to another. That is because to obtain IhQ in terms of IhG in
Eq. (11.55), we used the center of mass corollary. Nevertheless, there are occasions
when it is useful to be able to find IQ, the moment of inertia tensor about an arbitrary
point Q, given IP , the moment of inertia tensor about another arbitrary point P . You
can derive that expression in Problem 11.18.
11.4 Dynamics of Multibody Systems in
Three Dimensions
Section 9.6 discussed how to extend our results for the dynamics of rigid bodies and
particles to a collection of connected bodies and particles. Our conclusion was quite
simple: just break apart all connections, introduce constraint forces, and solve each
part separately. The constraint forces can be eliminated algebraically by properly
combining the equations of motion and constraints. In some cases we can avoid the
extra algebraic step of introducing constraint forces by using the parallel axis theorem
and the angular-momentum equation relative to a moving point. We also showed
that the total angular momentum and total energy of the system are conserved if the
total external moment is zero.
This idea is completely general and applies equally well to three-dimensional rigidbody problems. Simply break the components apart and use the equations of motion
introduced in this chapter or use the new three-dimensional version of the parallel axis
theorem (Eq. (11.56)). We can even use conservation of three-dimensional angular
momentum (or total energy) to study the behavior of the entire system. We specifically
highlight these ideas again in three dimensions because of a special class of connected
systems consisting of a rigid body with an attached spinning rotor (like the spinning
gyropendulum). Such a system is often used to stabilize a large moving mass, such
as a camera or ship, or to control the attitude of a rigid body, such as a spacecraft or
submarine. A rigid body with an attached spinning rotor is known as a gyrostat.It takes
advantage of what we discovered earlier in our qualitative analysis: when subject to
MULTIPARTICLE AND RIGID-BODY DYNAMICS IN THREE DIMENSIONS
503
an external moment, a system with significant angular momentum precesses rather
than spinning out of control. We illustrate how conservation of angular momentum
can be used in the example below. Tutorial 11.2 shows a more complex example using
gyroscopic stabilization.
Example 11.14 The Control Moment Gyroscope
The control moment gyroscope (CMG) is a device that utilizes conservation of total
angular momentum to control the rotation and orientation of a rigid body. It is
commonly used on large geosynchronous satellites. To a very good approximation
such a satellite is free of external torques and thus the total angular momentum
is conserved. Periodically, though, it may encounter an impulsive disturbance that
causes it to gain angular momentum. Rather than use a thruster (which applies an
external moment but requires fuel), the satellite applies an internal torque to one or
more large spinning disks that have significant angular momentum about their centers
of mass. By rotating the CMG, an angular-momentum component can be produced
in the direction opposite the satellite’s angular momentum. Since the total angular
momentum must be conserved (only internal moments are used to rotate the CMG),
the effect of the disturbance on the satellite’s rotation can be canceled out.
A fairly simple model of a satellite with a control moment gyroscope shows how
this works in principle. Let Ihs = Is . IωB be the angular momentum of the satellite, where Is is the moment of inertia tensor of the satellite and IωB is the angular velocity of a body-fixed frame B attached to the satellite. Likewise, Ihr is the
angular momentum of the control moment gyroscope’s rotor. The total angular momentum of the system is the sum of the satellite and gyroscope angular momenta and
is a constant of the motion:
I
hT = Is . IωB + Ihr = constant.
A CMG operates by reorienting the rotor spin in the body-frame so as to eliminate
the angular velocity of B with respect to I . The simplest model assumes the CMG
consists of a spinning rotor on a three-axis gimbal as in Example 10.7. If we introduce
a frame C fixed to the gyroscope rotor, then expressed as components in this frame the
rotor angular momentum is hc3. The objective is to set the CMG angular momentum
equal to the total angular momentum and thus force IωB = 0. Introducing matrix
notation and the direction-cosine matrix for the gimbals between frame C and frame
B yields
⎡ ⎤
0
B C⎣ ⎦
C
= IhT .
0
B
h C
These three equations can be solved for the 3-1-3 Euler angles (ψ, θ, φ)BC .
A real CMG system is considerably more complicated (Figure 11.17). Typically,
for reasons of redundancy and ease of assembly, CMGs are only mounted on a single
or double gimbal. It is thus common for up to four rotors to be employed on a satellite
and the solution for the gimbal motion involves more complicated control laws. An
excellent description of such a system can be found in Bryson (1994) and Wie (1998).
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CHAPTER ELEVEN
Figure 11.17 A control moment gyroscope of the type used on the International Space Station.
Image courtesy of L-3 Communications, Space, and Navigation.
11.5 Rotating the Moment of Inertia Tensor
Suppose you know the moment of inertia tensor of a rigid body B in some body
frame B1. How do you find the moment of inertia written in some other frame B2? The
answer is very similar to how we rotate the component magnitudes of a vector from
one frame to another. Recall from Eq. (10.38) that, if we have the direction-cosine
matrix describing the orientation of B1 in B2, then we can transform the component
magnitudes of a vector r written in B1 to those written in B2 using
[r]B2 = B2 C B1[r]B1.
A similar expression holds for the moment of inertia tensor. If we have the magnitudes of the moments and products of inertia in B1, then we can also find the moment
of inertia tensor in B2 by using the direction-cosine matrix B2 C B1. We just have to use
it twice:
[IG]B2 = B2 C B1[IG]B1
B
2
C B1 .
T
(11.65)
We have already seen one example where this expression would have been helpful.
Recall the discussion of principal axes in Section 11.2.4. We showed there the simplifications that arise in the equations of motion by choosing a principal-axes frame
but also noted that it sometimes is computationally (or experimentally) easier to find
MULTIPARTICLE AND RIGID-BODY DYNAMICS IN THREE DIMENSIONS
505
the moments and products of inertia in some other frame. To find the principal-axes
frame, you then need to find the direction-cosine matrix such that Eq. (11.65) results
in a diagonal matrix whose components are the principal moments of inertia. This is
an example of diagonalization, which you may remember from a linear algebra class.
For the matrix [IG]B2 to be diagonal, the direction-cosine matrix in Eq. (11.65) must
be the matrix of eigenvectors of [IG]B1 and the resulting principal moments of inertia
are the eigenvalues of [IG]B1. This is one of the most important reasons for discussing
rotation of the moment of inertia tensor—to transform to and from the principal axes.
One way to derive Eq. (11.65) is through the angular-momentum expression.
Suppose we have the angular velocity IωB1 of B1 in I . The angular momentum of
the rigid body expressed as components in B1 is
[IhG]B1 = [IG]B1[IωB1]B1.
(11.66)
Suppose we would instead like the angular momentum expressed as components
in frame B2. We can transform it by multiplying both sides of Eq. (11.66) by the
direction-cosine matrix:
[IhG]B2 = B2 C B1[IhG]B1 = B2 C B1[IG]B1[IωB1]B1.
(11.67)
The left-hand side is just [IhG]B2 , according to Eq. (10.38). For the right-hand side,
we also use Eq. (10.38), but on the angular velocity:
[IωB1]B2 = B2 C B1[IωB1]B1.
(11.68)
Using the orthogonality property of the direction cosine (i.e., that its inverse is also
its transpose), we can multiply both sides of Eq. (11.68) by (B2 C B1)T from the left
and then substitute the result into Eq. (11.67) to find
[IhG]B2 = B2 C B1[IG]B1
B
2
C B1 [IωB1]B2 .
T
Finally, we observe that IωB1 = IωB2 (since both frames are fixed to the rigid body),
which allows us to write the final form of the transformed angular momentum:
[IhG]B2 = B2 C B1[IG]B1
B
2
C B1 [IωB2 ]B2 .
T
(11.69)
We now compare Eq. (11.69) to the definition of the matrix form of the angular
momentum in terms of the moment of inertia matrix:
[IhG]B2 = [IG]B2 [IωB2 ]B2 .
Therefore it must be true that
[IG]B2 = B2 C B1[IG]B1
B
2
C B1 .
T
Example 11.15 A Long Thin Rod in a Gimbal
Consider the long thin rod connected to an outer gimbal by means of two rotations,
as shown in Figure 11.18. We wish to find the moment of inertia matrix for the rod in
frame A fixed to the outer gimbal. The rod has undergone a 3-1-3 rotation with respect
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CHAPTER ELEVEN
Figure 11.18 Gimbaled rod.
to the outer gimbal of ψ = 30◦, θ = 60◦, and φ = 0. The direction-cosine matrix AC Bp
for the principal-axes frame Bp fixed to the rod relative to the outer gimbal is thus
found from Example 10.7:
⎡ √3
⎤
1
⎤
⎡
0
◦
◦
2
2
sin 30
0
cos 30
⎢ 1 √3 √3 ⎥
A Bp
⎥
C = ⎣ − sin 30◦ cos 60◦ cos 30◦ cos 60◦
sin 60◦ ⎦ = ⎢
4
2 ⎦
4
⎣−
√
◦
◦
◦
◦
◦
sin 30 sin 60
− cos 30 sin 60 cos 60
1
3
− 43
4
2
Using Eq. (11.65) and the moment of inertia matrix found in Example 11.8, the
moment of inertia of the gimbaled rod in A is
√ ⎤
⎤ ⎡ √3
⎤ ⎡ ml 2
⎡ √3
1
3
0
− 41
0 0
2
2
2
4
12
√
√
√
⎥⎢ 1
⎥⎢
⎥
⎢ 1
3
3 ⎥⎢
3
ml 2
⎢
[IG]A = ⎢
− 43 ⎥
0⎥
4
2 ⎦⎣ 0
4
4
2
⎦
⎣
⎦.
⎣−
12
√
√
1
1
3
3
0
0 0
− 43
0
4
2 2
2 A B T
A B
= C p
=[IG ]Bp
=
C p
Multiplying these matrices gives the moment of inertia matrix of the rotated rod in
frame A:
⎤
⎡
1
0
0√
2
ml ⎢
1
− 43 ⎥
[IG]A =
⎦.
⎣0
4√
12
3
3
0 − 4
4
Why would we ever want to rotate the moment of inertia tensor to another frame?
In what situations would this be useful? As it turns out, there are many such situations.
MULTIPARTICLE AND RIGID-BODY DYNAMICS IN THREE DIMENSIONS
h
507
β
l1
l2
θ1
h
–h
2
(a)
C
A
O
a2
B
c2
O″
b2
G2
G1
c1
b1
a1
O′
(b)
Figure 11.19 A model of a crane. (a) Crane schematic. (b) Reference frames.
A common reason for using Eq. (11.65) is to find the moment of inertia tensor of a
rigid body composed of many connected rigid bodies of different shapes. The easiest
way to find the composite moment of inertia tensor is to find the inertia tensor of each
individual rigid body in its own principal-axes frame, rotate it into the body frame of
the composite rigid body, and then use the parallel axis theorem to sum the constituent
moment of inertia matrices, as demonstrated in the next example.
Example 11.16 The Moment of Inertia of a Crane
A crane, shown schematically in Figure 11.19, is composed of a cab (modeled as a
cube of mass M) and two rigid links that can rotate freely in a plane (modeled as
long thin rods of masses m1 and m2). We wish to find an expression for the moment
of inertia tensor IO of the whole system about the center of the cab (point O) and
expressed as components in frame A = (O, a1, a2, a3) fixed to the cab. This might be
necessary, for instance, to find the torque needed to rotate the cab around the a2 axis.
To calculate IO , we find the moment of inertia tensor for each of the three components
(cab and two arms), express each as a matrix in A, and sum them.
Example 11.6 determined the moment of inertia of a constant-density cube. Using
this result, we write the moment of inertia of the cab about point O in frame A as
Mh2
2 h2
cab
I,
]A = M I =
[IO
3 4
6
where M is the mass of the cab and I is the identity matrix.
(11.70)
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CHAPTER ELEVEN
Example 11.8 determined the moment of inertia of a long thin rod aligned with the
b3 axis. Here we have a long thin rod aligned with the b1 axis. The moment of inertia
of the first arm about its center of mass (G1, located halfway along the arm) in frame
B is
⎤
⎡
0 0 0
m
1
[IG
] = 1 ⎣ 0 l12 0 ⎦ ,
1 B
12
0 0 l12 B
where m1 is the mass of the first arm. Following Eq. (11.65),
1
1
]B
]A = A C B [IG
[IG
1
1
A
CB
T
,
where
⎤
cos θ1 − sin θ1 0
A B
C = ⎣ sin θ1 cos θ1 0 ⎦ .
0
0
1
⎡
1 ] by noting that
We can use the parallel axis theorem (Eq. (11.57)) to calculate [IO
l1
h
h l1
l
+ cos θ1 a1 − 1 sin θ1a2 .
rO/G1 = − b1 − a1 = −
2
2
2
2
2
Putting this all together, we have
1
1
[IO
]A = A C B [IG
]B
⎡
=
m1 ⎢
⎣
12
A
1
CB
T
+ m1 rO/G12 I − [rO/G1]A [rO/G1]TA
(11.71)
4l12 sin2 θ1
I12
0
I21
3h2 + 4l12 cos2 θ1 + 6hl1 cos θ1
0
0
0
3h2 + 4l12 + 6hl1 cos θ1
⎤
⎥
⎦,
A
θ1 − 2l12
sin 2θ1.
where I12 = I21 = −3hl1 sin
The moment of inertia of the second arm in frame C about its center of mass G2
is the same as that of arm 1 in frame B about G1 (except with l2 and m2 instead of l1
and m1), so we perform the same calculations as before, with
⎤⎡
⎤
⎡
cos β
sin β 0
cos θ1 − sin θ1 0
A C
C = A C BB C C = ⎣ sin θ1 cos θ1 0 ⎦ ⎣ − sin β cos β 0 ⎦
0
0
1
0
0
1
⎤
⎡
cos θ2
sin θ2 0
= ⎣ − sin θ2 cos θ2 0 ⎦ ,
0
0
1
where θ2 = β − θ1. It is important to note that in Figure 11.19, β is defined as
increasing in the opposite direction from θ1, making it a left-handed, rather than
a right-handed, rotation. (There’s nothing wrong with doing this, but one must be
careful to keep track of the direction of each rotation—which is why the directioncosine matrix B C C has opposite signs from A C B on the sine terms.) The angle θ2
increases in the same direction as θ1.
MULTIPARTICLE AND RIGID-BODY DYNAMICS IN THREE DIMENSIONS
509
Using A C B and A C C , we can write
l
h
rO/G2 = − 2 a1 − l1b1 − a1
2
2
h
l
l
=−
+ l1 cos θ1 + 2 cos θ2 a1 + −l1 sin θ1 + 2 sin θ2 a2
2
2
2
T
2
2 ] = A C C [I2 ] A C C T + m
and thus, using [IO
2 rO/G2 I − [rO/G2 ]A [rO/G2 ]A ,
A
G
2
⎡
I11 I12
m
2 ⎣
2
[IO ]A =
I21 I22
12
0
0
⎤
0
0 ⎦ ,
I33 A
(11.72)
where
I11 = 12l12 sin2 θ1 − 12l1l2 sin θ1 sin θ2 + 4l22 sin2 θ2
I12 = I21 = 2l22 sin 2θ2 + 3hl2 sin θ2 − 6l1l2 sin(θ1 − θ2) − 6hl1 sin θ1 − 6l12 sin 2θ1
2
I22 = 3 h + 2l1 cos θ1 + 4l22 cos2 θ2 + 6l2(h + 2l1 cos θ1) cos θ2
I33 = 3h2 + 12l1 h cos θ1 + l1 + 4l22 + 6hl2 cos θ2 + 12l1l2 cos β .
The total moment of inertia about O in frame A is the sum of the inertia matrices in
Eqs. (11.70)–(11.72):
⎤
⎡ I12
0
I
1 ⎣ 11
2
1
cab
[IO ]A = [IO ]A + [IO ]A + [IO ]A =
I21
I22
0 ⎦ ,
12
0
0 I33
A
where
= 2Mh2 + 4(m1 + 3m2)l12 sin2 θ1 − 12l1l2m2 sin θ1 sin θ2 + 4m2l22 sin2 θ2
I11
= I21
= −2(m1 + 3m2)l12 sin 2θ1 + 2m2l22 sin 2θ2
I12
− 3 hl1(m1 + 2m2) sin θ1 + l2m2(2l1 sin(θ1 − θ2) − h sin θ2)
I22
= 2Mh2 + 3(m1 + m2)h2 + 4(m1 + 3m2)l12 cos2 θ1
+ 6(m1 + 2m2)hl1 cos θ1 + 6m2l2(h + 2l1 cos θ1) cos θ2 + 4m2l22 cos2 θ2
I33
= 2Mh2 + 3(m1 + m2)h2 + 4(m1 + 3m2)l12 + 4m2l22 + 6(m1 + 2m2)hl1 cos θ1
+ 6m2l2(h cos θ2 + 2l1 cos β) .
11.6 Angular Impulse in Three Dimensions
Section 9.3.4 discussed how to treat a rigid body after an impulsive moment MG(t1, t2)
is applied. The solution to the vector equation of motion for the angular momentum
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CHAPTER ELEVEN
in Eq. (11.3) does not change for an impulsive moment because we did not use the
planar assumption in Section 9.3.4. We have
I
hG(t2) = IhG(t1) + MG(t1, t2).
(11.73)
While Eq. (11.73) is true, it is not as useful for solving problems as one might think
because we usually are looking for the angular velocity at t = t2 so we can integrate
to find the orientation of the rigid body (or describe the trajectory after the impulse).
For a planar rigid body, this was accomplished by dividing by the moment of inertia.
It is slightly more complicated in three dimensions but not excessively so. Here it is
easiest to turn to matrix notation. We rewrite Eq. (11.73), using Eq. (11.23), to obtain
[IG]B [IωB (t2)]B = [IG]B [IωB (t1)]B + [MG(t1, t2)]B .
(11.74)
Eq. (11.74) allows us to solve for the instantaneous change in the angular velocity due
to an impulsive moment by inverting the moment of inertia matrix and multiplying
from the left:
[IωB (t2)]B = [IωB (t1)]B + [IG]−1
B [MG (t1, t2 )]B .
Just as in the planar case, the application of an impulsive moment instantaneously
changes the angular velocity. If the moment impulse is a result of a force (rather than
a pure torque), then the velocity of the center of mass will also be changed, just as in
Example 9.11. We are able to find the angular velocity at t2 because the moment of
inertia matrix is always invertible.
11.7 Work and Energy of a Rigid Body in Three Dimensions
Our final task in this chapter is to extend the discussion in Chapters 7 and 9 of work
and energy for planar multiparticle systems and rigid bodies to three dimensions.
Although all the basic principles still hold, in some cases we need to use a bit of care,
just as when discussing the angular momentum.
11.7.1 Kinetic Energy of a Rigid Body in Three Dimensions
The earlier derivation in Chapters 7 and 9 of the separation of motion of the center of
mass and motion about the center of mass of a collection of particles was completely
general. The result that the kinetic energy can be written as the sum of the kinetic
energy due to motion of the center of mass and the kinetic energy due to motion
about the center of mass applies unchanged in three dimensions:
TO = TG/O + TG.
As before, TG/O is the translational kinetic energy of the center of mass:
1
mGIvG/O 2.
TG/O =
2
MULTIPARTICLE AND RIGID-BODY DYNAMICS IN THREE DIMENSIONS
511
The derivation of TG—the kinetic energy due to motion about the center of mass—
in terms of the moment of inertia tensor for a rigid body requires a bit of care. By
definition, for a system of N particles we have
TG =
N
N
1 I
1
mi vi/G . Ivi/G =
mi Ivi/G2.
2 i=1
2 i=1
Applying the transport equation to Ivi/G and making use of the fact that we are
studying a rigid collection of N particles yields
TG =
N
N
1
1
mi (IωB × ri/G) . (IωB × ri/G).
mi Bvi/G +IωB × ri/G2 =
2 i=1
2 i=1
=0
Next we apply two cross-product identities from Appendix B—the scalar triple
product and the vector triple product. Sequential application of these two identities
results in
TG =
=
N
1 I B.
mi ω (ri/G × (IωB × ri/G))
2 i=1
N
1I B . ω
mi (ri/G . ri/G)IωB − ri/G(ri/G . IωB ) ,
2
i=1
(11.75)
where we pulled the angular velocity out of the sum over i because it does not depend
on i.
Note the similarity of Eq. (11.75) to Eq. (11.14). As in Section 11.2.2, we now
invoke tensor analysis to complete the derivation. Factoring out the angular velocity
to the right yields
N
I B
1I B . .
TG = ω
mi (ri/G ri/G)U − (ri/G ⊗ ri/G) . ω ,
2
i=1
=IG
where U is the unity tensor and IG is the moment of inertia tensor about the center of
mass. The result is that the rotational kinetic energy of a general rigid body about its
center of mass has the compact form
TG =
1I B . . I B
ω IG ω .
2
In matrix notation, the rotational kinetic energy is
1
TG = [IωB ]TB [IG]B [IωB ]B .
2
Although not quite as simple as for the planar case, we still have an elegant
expression for the kinetic energy of rotation that depends only on the angular velocity
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CHAPTER ELEVEN
of the rigid body and its mass properties. For an arbitrary body frame, the resulting
expression for TG can be rather cumbersome. However, as long as we work in principal
axes, the following expression:
1
TG = (I1ω12 + I2ω22 + I3ω32)
(11.76)
2
holds, where ω1, ω2, and ω3 are the coefficients of IωB expressed as components in B.
Our result for TG also shows that, as in Eq. (9.40), we can write the kinetic energy
of rotation in terms of the angular momentum in three dimensions:
TG =
1I . I B
hG ω .
2
Finally, although we will not rederive it here (you can try it in Problem 11.4), we
can write the total kinetic energy in terms of rotation relative to an arbitrary point Q
on the rigid body exactly as in Eq. (9.41):
TO = TQ/O + TQ − mGIvQ/O . (IωB × rQ/G),
(11.77)
where the kinetic energy due to rotation relative to Q is
TQ =
1I B . . I B
ω IQ ω
2
and the moment of inertia tensor relative to Q is given by the parallel axis theorem
in Eq. (11.56).
11.7.2 Three-Dimensional Rigid-Body Work, Potential Energy,
and Total Energy
Our next task is to generalize the work–kinetic-energy formulas for three-dimensional
rigid bodies. Fortunately, the task is again greatly simplified by the fact that the
development in Chapter 9 was completely general; it never actually invoked the planar
restriction. Thus we can simply jump right to the work–kinetic-energy formula in
Eq. (9.42),
W = TG/O + TG,
where W = W ({ri/O ; γi }N
i=1) refers to the total external work on all of the particles along their three-dimensional trajectories. For a rigid body, we again separate the external work into the work due to motion of the center of mass WG/O =
WG/O (rG/O , θ, k; γG, γθ,k ) and the work due to rotation about the center of mass
WG = WG(rG/O , θ, k; γG, γθ,k ). The former satisfies the three-dimensional version
of Eq. (9.45),
WG/O = TG/O ,
whereas the latter satisfies the three-dimensional version of Eq. (9.47):
WG = TG.
Remember, though, that for the most general cases, the work WG can depend on the
trajectory of the center of mass embedded in WG/O and vice versa.
MULTIPARTICLE AND RIGID-BODY DYNAMICS IN THREE DIMENSIONS
513
As in Chapter 9, γG represents the trajectory of the center of mass. However, we
have had to modify our notation a bit for motion about the center of mass. When we
were considering a planar rigid body, the rotational trajectory was entirely described
by the single angle θ. Thus the work was the integral of the moment times the angular
displacement over the trajectory. Here, things are somewhat more complicated, as
the rigid body can rotate arbitrarily in three dimensions. Rather than parameterize the
work by the three Euler angles (which vary, depending on the set of angles chosen),
we have selected to parameterize it by the unit vector k along the Euler axis and the
instantaneous rotation angle θ about that axis. Thus the instantaneous work is the
product of the moment about the Euler axis times the rotation angle θ. As shown
below, this choice is sensible, as the work is more often computed by means of the
velocity form, integrating the inner product of the external moment and the angular
velocity. It also highlights the fact that only moments along the Euler axis (that is, in
the same direction as the angular velocity) do work. Moments perpendicular to that
axis do no work, just as forces perpendicular to the translational direction of motion
(like constraint forces) do no work. As shown in Section 10.5, the angular velocity is
along the instantaneous axis of rotation; it is always aligned with the Euler axis. Thus
we describe the rotational trajectory by the parameter γθ,k .
As in Section 9.5.2, we write the work due to motion about the center of mass in
terms of the applied moments:
t2
WG(rG/O , θ, k; γG, γθ,k ) =
MG . IωB dt.
t1
The corresponding power is
PG = MG . IωB .
As in Chapter 9, our final step is to replace the external work by a potential energy
when the corresponding forces or torques are conservative. There was no restriction
made in Section 9.5.3 to planar motion other than in computing the pure torque from
its corresponding potential. Thus the total energy is still defined as
TG/O (t) + TG(t) + UO (t),
EO (t) =
where the total potential energy is
UO (t) =
N
Ui (ri/O (t)).
i=1
Likewise, if we have a rigid body B rather than a collection of N particles, the
external potential is given by Eq. (9.48):
Udm/O (rdm/O ) dm.
UO (t) =
B
The result is that the same work-energy formula with conservation of energy holds
in three dimensions:
(nc)
EO (t2) = EO (t1) + WG/O
+ WG(nc).
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CHAPTER ELEVEN
Example 11.17 Torque-Free Motion, Revisited
Example 11.10 examined the torque-free motion of a spinning rigid body, but restricted to small deviations of the angular velocity from the nominal spin. For spin
about the major or minor axis of inertia, the angular velocity rotated about the inertia
axis; spin about the intermediate axis was unstable. In this example we take a different approach—using energy conservation—to find the complete set of trajectories
corresponding to the possible motion of the angular velocity in the body frame.
We begin by noting that, if the total external force and total external moment are
zero, then the total energy of the rigid body is equal to its kinetic energy and the
kinetic energy is a constant of the motion. Thus in a principal-axes body frame, the
kinetic energy about the center of mass is given by Eq. (11.76):
1
TG = (I1ω12 + I2ω22 + I3ω32).
2
Dividing through by TG yields
ω12
2TG/I1
+
ω22
2TG/I2
+
ω32
2TG/I3
= 1.
(11.78)
for an ellipsoid in the body frame with semi-axes
is the equation
Eq. (11.78)
2TG/I1, 2TG/I2, and 2TG/I3. Recall that the Cartesian equation for an ellipse
in two dimensions is
x2 y2
+ 2 = 1,
a2
b
where a is the semi-major axis and b is the semi-minor axis. Eq. (11.78) simply adds
a third dimension. We call this form the inertia ellipsoid, as its shape is determined by
the three principal moments of inertia. Typically, the axes are numbered such that I3 >
I2 > I1. Eq. (11.78) states that, for a given set of initial conditions on the angular velocity, which establishes the kinetic energy TG, the tip of the angular velocity will move
in the body frame in such a way as to always be on the surface of the inertia ellipsoid.
However, we have another constraint. Since the total external moment is zero, the
angular momentum is also a constant of the motion; it is fixed in both magnitude and
direction (in the inertial frame). Expressing the inertia tensor and angular velocity as
components in a principal-axes body frame, we can find the squared magnitude of the
angular momentum using Eq. (11.15) or Eq. (11.23):
IhG2 = h2G = I12ω12 + I22ω22 + I32ω32.
After a manipulation similar to that done on Eq. (11.78) we find
ω12
h2G/I12
+
ω22
h2G/I22
+
ω32
h2G/I32
= 1.
(11.79)
This is the equation for a second ellipsoid, whose size depends upon the angular momentum, that the angular velocity must also trace. Thus the path of the
angular velocity in the body frame is given by the curve representing the intersection
of the first ellipsoid in Eq. (11.78) and the second ellipsoid in Eq. (11.79). These curves
are called polhodes. Figure 11.20 shows various polhode curves on the surface of the
first ellipsoid. Note that the curves collapse to a point at each of the three principal
MULTIPARTICLE AND RIGID-BODY DYNAMICS IN THREE DIMENSIONS
515
I3
I1
I2
Figure 11.20 Inertia ellipsoid of a rigid body showing the polhode curves that trace the path
of the angular velocity for torque-free motion.
axes, representing spin about each of these axes, as discussed in Example 11.10. For
slight deviations from the axes of maximum and minimum moments of inertia (I3
and I1, respectively) the angular velocity encircles the principal axis, as discussed
in Example 11.10. For slight deviations from the intermediate axis, however, the
angular velocity travels away from the axis along one of two unstable trajectories
called separatrices. This graphical approach gives a complete picture of all curves
traced by the angular velocity, not just those for small motions about a spin axis.
We conclude this example with an important observation. Suppose that the rigid
body is spinning about a single principal axis with moment of inertia Ii , where
i = 1, 2, or 3. Then Eqs. (11.78) and (11.79) can be combined to find
TG =
h2G
2Ii
.
This equation tells us that, for torque-free motion where hG is constant, the kinetic
energy is a maximum for spin about the minimum axis of inertia and a minimum for
spin about the maximum axis of inertia. Suppose that there is some nonconservative
work in the rigid body, causing dissipation. This work will always cause the angular
velocity to change from spin about the minimum to that about the maximum axis
to minimize the energy. This is admittedly a qualitative argument and ignores the
specific mechanisms for such motion, but it was observed in the early days of the
space program, when satellites that were set spinning about the minimum axis of
inertia began to tumble out of control due to internal dissipation in antenna structures.
So despite the theoretical stability of the polhode paths about the minimum axis, such
spin is often unstable anyway!
11.8 Tutorials
Tutorial 11.1 Torque-Free Motion of a Symmetric Rigid Body
Example 11.10 used Euler’s equations to approximately solve for the motion of the
angular velocity in the body frame of a torque-free spinning rigid body. Example 11.17
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CHAPTER ELEVEN
used energy to find the set of permissible trajectories of the angular velocity in the
body frame. This tutorial returns again to the torque-free rigid body and solves for the
trajectory of the body in the inertial frame. To simplify, we only examine the special
case of a symmetric rigid body. In this case I1 = I2, and there is no intermediate
axis. Recall that spin about the I3-axis (maximum or minimum moment of inertia) is
stable. If you’ve ever seen someone juggling pins, you have seen torque-free motion
of a spinning symmetric rigid body. (In that case, the rotation is about the major axis.)
We begin again with Euler’s equations, with ω3 = ωs the (large) spin about the
3-axis. From Example 11.10 we have
ω̈1 + ωn2 ω1 = 0
ω̈2 + ωn2 ω2 = 0,
where
ωn2 =
(IO − J )2 2
ωs ,
IO2
(11.80)
I1 = I2 = IO , and J = I3. Thus the variation of each angular velocity component is
simple harmonic with solution
ω1 = −ω0 cos ωnt
ω2 = ω0 sin ωnt,
where we arbitrarily chose ω1(t0) = ω0 and ω2(t0) = 0. It is possible to verify that the
oscillatory motion of ω1 and ω2 is exactly 90◦ out of phase by substituting back into
Eqs. (11.40) and (11.41).
How does the motion of the body in the inertial frame deviate from simple spin due
to the small ω1 and ω2 perturbations? Examining the angular momentum will answer
this question. Since there are no torques on the rigid body, the angular momentum IhG
has constant magnitude and direction. Using the three-dimensional expression for the
angular momentum of a rigid body in principal axes (Eq. (11.15) or Eq. (11.23)), we
can write the components of IhG in the body frame:
I
hG = IO ω0 cos ωntb1 − IO ω0 sin ωntb2 + J ωs b3.
(11.81)
The b3 symmetry axis of the rigid body cones about the inertial direction defined by
Ih . We find the angle θ between the body symmetry axis and the angular momentum
G
by dotting IhG in Eq. (11.81) with b3 and dividing by IhG:
cos θ =
Ih
.
G b3
I
hG
=
J ωs
.
(11.82)
IO2 ω02 + J 2ωs2
We call this coning motion free-body precession; the angle θ is the nutation angle.
This effect is also often referred to as spin stabilization because spinning the rigid
body stabilizes the orientation of its spin axis against impulsive disturbances. Suppose
we want the symmetry axis of the body to point in a particular inertial direction.
Inevitably, it will be perturbed by a small impulse. If the body is not spinning, it will
develop a small angular velocity in response to the impulse and drift away. However,
MULTIPARTICLE AND RIGID-BODY DYNAMICS IN THREE DIMENSIONS
517
by spinning it, the small angular velocity due to the impulse results in a precession
about the desired inertial direction at an angle given by Eq. (11.82). For this reason
all satellites are spun before being transferred from a low orbit to a high one (see
Tutorial 4.3). Any small alignment errors in the applied impulse induce a small coning
motion rather than causing the satellite to tumble out of control.
We finish this tutorial by examining a 3-2-3 Euler-angle description of the rigid
body. Since the angular momentum defines a fixed direction in absolute space, we
can use it to define one of the axes of the inertial frame. We thus let e3 = IhG/IhG.
Eq. (11.82) determines the fixed tilt angle between IhG (and thus e3) and the body b3axis, implying that θ̇ = 0.5 The kinematic equations of rotation in Eq. (10.48) become
ω1 = −ω0 cos ωnt = −ψ̇ sin θ cos φ
(11.83)
ω2 = ω0 sin ωnt = ψ̇ sin θ sin φ
(11.84)
ω3 = ωs = φ̇ + ψ̇ cos θ.
(11.85)
Squaring and adding Eqs. (11.83) and (11.84) leaves
ω02 = ψ̇ 2 sin2 θ.
Since θ and ω0 are fixed, ψ̇ is a constant and represents the precession rate of the b3
axis about the angular-momentum direction.
We now use the transformation array from Eq. (10.33) to solve for the motion of
the body frame and the angular velocity in the inertial frame. First, we can see again
that the body symmetry axis cones about the inertial e3 direction (aligned with the
angular momentum) by writing b3 as components in I using the third row of the
transformation array,
b3 = cos ψ sin θe1 + sin ψ sin θe2 + cos θe3.
(11.86)
Since θ and ψ̇ are constant, the e1 and e2 components vary sinusoidally 90◦ out of
phase; b3 rotates or cones about e3 at a precession rate ψ̇.
To find the precession rate, we can use the transformation array in Eq. (10.33) to
write the angular momentum, I hG = IGe3, in the body frame,
I
hG = hGe3 = −hG sin θ cos φb1 + hG sin θ sin φb2 + hG cos θb3.
Since I hG = IG . I ωB = IO ω1b1 + IO ω2b2 + J ωs b3, we can equate components and
use the kinematical equations in Eqs. (11.83)–(11.85) to find that ψ̇ = hG/IO . The
body symmetry axis thus precesses in a positive direction about e3 at a rate given by
the magnitude of the angular momentum.
We likewise use the transformation array in Eq. (10.33) to write the angular
velocity, I ωB = ω1b1 + ω2b2 + ω3b3, as components in the inertial frame,
I
ωB = (ωs sin θ − ω0 cos θ) cos ψe1 + (ωs sin θ − ω0 cos θ) sin ψe2
+ (ωs cos θ + ω0 sin θ)e3,
5 Unlike
that for the symmetric rigid body undergoing torque, the nutation angle of the torque-free rigid
body does not vary with time.
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CHAPTER ELEVEN
where we also used the fact that ω3 = ωs and substituted for ω1 and ω2 from
Eqs. (11.83) and (11.84), using the fact that ψ̇ > 0. Comparing to Eq. (11.86) shows
that the angular velocity and the symmetry axis are in the same plane as the angular
momentum and rotate around it either in phase or 180◦ out of phase, depending upon
the sign of the constant amplitudes of the e1 and e2 components, ωs sin θ − ω0 cos θ.
The sign of this term can be found in terms of the moments of inertia by substituting
for cos θ (from Eq. (11.82)) and sin θ. We find an expression for sin θ by taking the
cross product of b3 with the angular momentum and using the definition of the cross
product (Definition B.3),
sin θ =
IO ω0
b3 × I hG
=
.
I
hG IO2 ω02 + J 2ωs2
The amplitude of the e1 and e2 components of the angular velocity is thus
(I − J )ω0ωs
ωs sin θ − ω0 cos θ = O
.
IO2 ω02 + J 2ωs2
Last, we can find an expression for the relative spin, φ̇, from Eq. (11.85) and the
b3 component of the angular momentum,
I −J
I O − J hG
cos θ = O
ωs ,
φ̇ =
IO
J
IO
where we also used ψ̇ = hG/J .
If IO > J , then (ωs sin θ − ω0 cos θ) > 0 and φ̇ > 0, implying that b3 and I ωB are
on the same side of I hG and rotate around it in the same direction as the relative spin.
This motion is called direct or prograde precession and the rigid body with IO > J
is called prolate. It is long and thin, like a pencil.
If IO < J , then (ωs sin θ − ω0 cos θ) < 0 and φ̇ < 0, implying that b3 and I ωB are
on opposite sides of I hG and rotate about it in the opposite direction of the relative
spin. This motion is called retrograde precession and the rigid body with IO < J is
called oblate. It is flat and broad, like a pancake.
The torque-free precession of a rigid body is often geometrically described by two
right circular cones rolling on each other, as shown in Figure 11.21. The space cone
surrounds the fixed angular momentum in absolute space and is defined by the angle
between IhG and IωB . The body cone surrounds the body-frame 3-axis and is defined
by the angle between IωB and b3. (The angular velocity is directed along the line
of contact between the two cones.) For a prolate rigid body, precession of b3 about
Ih is described by the body cone rolling around the outside of the space cone, as in
G
Figure 11.21a. For an oblate rigid body, the inside of the body cone rolls around the
space cone, as in Figure 11.21b. Note the different arrangements of the three vectors,
Ih , IωB , and b , in each case.
3
G
Tutorial 11.2 Ship Stabilization
At one time, ship designers considered using a large gyroscope mounted to the hull
to damp out the rocking motion. While this approach proved less practical for very
MULTIPARTICLE AND RIGID-BODY DYNAMICS IN THREE DIMENSIONS
I
I
e3
e3
θ
θ
I
519
I
hG
hG
B
b3
B
b3
Space cone
I
ωB
Space cone
Body cone
G
I
ωB
Body cone
G
(a)
(b)
Figure 11.21 Space and body cones for the precession of a rigid body. (a) Prolate case IO > J .
(b) Oblate case IO < J .
(a)
(b)
Figure 11.22 Example of a commercially available ship-stabilizing gyroscope for private
yachts. (a) The gyroscope and gimbal mounted to the hull. (b) Cutaway view of the rotor.
Images courtesy of Seakeeper, Inc.
large ships because of the large torques imposed on the superstructure, such a system
is available for small yachts (see Figure 11.22). It is an excellent example of using a
gyrostat to stabilize a vehicle. In this tutorial, we explore how it works.
Figure 11.23 shows the front view of a ship in inertial frame I = (O, e1, e2, e3),
whose orientation is described by the single Euler angle φ about the e1 axis. Attached
to the hull of the ship is a spinning disk (the rotor) that is allowed to pitch forward
and backward about pivots attached to the rotor and the ship. The angle that the plane
of the rotor makes with the hull is given by coordinate .
We consider the simple situation where the ship is traveling straight ahead and
rolling back and forth because of wave action. We solve this problem by separating
the ship and the rotor and including constraint torques at the pivot. Figure 11.24 shows
the two body frames, frame B = (G, b1, b2, b3) attached to the ship at its center of
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CHAPTER ELEVEN
ϕ
I
e3
hc3
O
e2
e1
Figure 11.23 Front view of a rocking ship with gyro stabilization.
c3
B
C
b3
c1
G
ϕ
b2
O′
b1
b1
c2, b2
(a)
(b)
Figure 11.24 Ship and rotor body frames and coordinates. (a) Ship frame (front view).
(b) Rotor frame (side view).
mass, with b1 = e1, and a frame C = (O , c1, c2, c3) attached to the rotor, with c2 = b2.
The transformation table between the ship’s body frame and the inertial frame is
e1
e2
e3
b1
b2
1
0
0
0
cos φ
sin φ
b3
0
− sin φ
cos φ .
The angle that the rotor makes with the ship’s hull is given by angle
b2 = c2 axis, resulting in the transformation table
b1
b2
b3
c1
c2
cos
0
sin
0
1
0
about the
c3
− sin
0
cos
.
Our goal is to find equations of motion for φ and . To do so, we split apart the two
rigid bodies at the pivot and introduce the constraint moment acting between them to
keep the rotor and ship aligned, given by pure torques about the b1 and b3 axes, as
shown in the free-body diagrams in Figure 11.25. The moment on the ship is Ms =
τ1b1 + τ3b3, and the moment on the rotor is equal and opposite: Mr = −τ1b1 − τ3b3.
We find the rigid-body equations for the ship and rotor separately, starting with the
ship. The angular velocity of the ship in the inertial frame is
MULTIPARTICLE AND RIGID-BODY DYNAMICS IN THREE DIMENSIONS
521
(τ1 sin – τ3 cos )c3
τ3b3
–(τ1 cos + τ3 sin )c1
G
–(D + kr)c2
τ1b1 – ksϕb1
(a)
(b)
Figure 11.25 (a) Ship and (b) rotor moment diagrams.
I B
ω = φ̇b1.
If we assume the ship has a moment of inertia about b1 of I1, then the angular
momentum of the ship about its center of mass is
I
hG = I1φ̇b1.
We now use Eq. (11.26) or Eq. (11.27) and the moment diagram in Figure 11.25a
to find the equation of motion of the ship. Since the angular velocity and angular
momentum are aligned, the cross product from the transport equation is zero, and we
have simply
I1φ̈ = −ks φ + τ1,
(11.87)
where we have introduced a restoring moment due to the water with spring constant ks .
We next find the equations of motion for the rotor. The angular velocity of frame
C in B is
B C
ω = ˙ c2 .
Combining this expression with the angular velocity of B in I by means of the
addition property and using the transformation array between B and C gives the
angular velocity of the rotor in absolute space:
I C
ω = φ̇ cos c1 + ˙ c2 − φ̇ sin c3.
(11.88)
We assume that the symmetric rotor has a large inertia Ir about its spin axis and a
smaller moment of inertia J about each of the transverse axes, giving it a moment of
inertia tensor:
⎤
⎡
J 0 0
[IO ]C = ⎣ 0 J 0 ⎦ .
0 0 Ir C
Multiplying the moment of inertia tensor and the angular velocity yields the angular
momentum of the rotor as components in frame C,
I
hO = IO . IωC + hc3 = J φ̇ cos c1 + J ˙ c2 + (h − Ir φ̇ sin )c3, (11.89)
where we have included the constant spin angular momentum h about the c3 axis.
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CHAPTER ELEVEN
We now use Eq. (11.26) or Eq. (11.27) to find the equations of motion, but this time
in the C frame. The moment on the rotor from the moment diagram in Figure 11.25b
expressed as components in frame C is
Mr = −(τ1 cos + τ3 sin )c1 + (τ1 sin − τ3 cos )c3 − (D ˙ + kr )c2,
where we have introduced a restoring torque kr and a viscous damper D on the pivot
between the rotor and the ship. In most systems the torque and damping would be
supplied by a feedback-control system on the rotor, but it also works with passive
components. (Note that we did not include the equal-and-opposite pivot forces when
examining the equations of motion of the ship, as we only consider motion of the ship
about the b1 = e1 direction.)
Taking the inertial derivative of Eq. (11.89) using the transport equation and the
angular velocity in Eq. (11.88) and setting the result equal to the moment gives three
scalar equations:
J φ̈ cos + ˙ (h − Ir φ̇ sin ) = −τ1 cos − τ3 sin
(11.90)
J ¨ − J φ̇ 2 sin cos − φ̇ cos (h − Ir φ̇ sin ) + D ˙ + kr = 0
(11.91)
Ir φ̈ sin + Ir φ̇ ˙ cos = −τ1 sin + τ3 cos .
(11.92)
We can reduce these to two equations for φ and by multiplying Eq. (11.90) by
cos and Eq. (11.92) by sin , adding, and solving for τ1. Then τ1 is substituted into
Eq. (11.87) to obtain the equation of motion for φ̈. Combined with Eq. (11.91), this
result yields our two equations of motion. Let ωs2 = ks /I1 be the natural frequency
of the ship’s rocking, ωr2 = kr /J be the natural frequency of the rotor oscillation,
2ζr ωr = D/J be the rotor damping, K1 = h/I1, and K2 = h/J . We have
J cos2 + Ir sin2
1+
φ̈ + ωs2φ + K1 ˙ cos = 0
I1
I φ̇ sin
¨ + 2ζr ωr ˙ + ωr2 − φ̇ 2 sin cos − cos φ̇ K2 − r
= 0.
J
Although these equations are fairly complicated, they are easily integrated to
show that adding the gyroscope stabilizes the ship. Figure 11.26 shows a numerical
simulation for typical parameter values and three different damping constants. For
these simulations, we chose a natural frequency for the ship’s rocking of 1 Hz and
for the rotor of 10 Hz. As the damping constant in the rotor attachment increases, the
ship’s rocking damps out more quickly.
Analysis of the ship stabilization system can be simplified by linearization. If we
assume that both the ship and the rotor motions are small ( , ˙ , φ, φ̇ 1), then we can
drop all second-order terms using the small-angle approximation to find the following
linear equations of motion for the ship and rotor:
φ̈ + ω̄s2φ + K̄1 ˙ = 0
¨ + 2ζr ωr ˙ + ωr2 − K2φ̇ = 0,
where ω̄s2 = ωs2/(1 + J /I1) and K̄1 = K1/(1 + J /I1). Chapter 12 explains how solutions of coupled linear equations of motion are always harmonic, as in Figure 11.26.
20
20
15
15
10
10
5
5
Angle (°)
Angle (°)
MULTIPARTICLE AND RIGID-BODY DYNAMICS IN THREE DIMENSIONS
0
–5
0
–5
–10
–10
–15
–15
–20
0
2
4
6
8 10 12 14 16 18 20
t (s)
523
–20
0
2
4
6
(a)
8 10 12 14 16 18 20
t (s)
(b)
20
15
Angle (°)
10
5
0
–5
–10
–15
–20
0
2
4
6
8 10 12 14 16 18 20
t (s)
(c)
Figure 11.26 Roll of the ship (solid curves) and pitch of the rotor (dashed curves) for three
different damping constants. Here ωs = 1 Hz, ωr = 10 Hz, K1 = 50 s−1, and K2 = 100 s−1.
(a) ζr = 0. (b) ζr = 0.4. (c) ζr = 0.8.
11.9 Key Ideas
.
The translational dynamics of the center of mass of a multiparticle system in
three dimensions are
(ext)
FG
= mGIaG/O .
.
The total angular momentum of a three-dimensional N-particle system about
point O has the dynamics
Id
dt
.
N
(ext)
.
MO
hO =
ri/O × Fi(ext) =
I
i=1
The total angular momentum of a three-dimensional multiparticle system about
point O separates into the angular momentum of the center of mass G about O
and the angular momentum about G:
I
hO = IhG/O + IhG,
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CHAPTER ELEVEN
where
hG/O =
rG/O × mGIvG/O
I
and
I
hG =
N
mi ri/G × Ivi/G.
i=1
.
The rotational dynamics of a three-dimensional rigid body about its center of
mass G are
Bd I
hG + IωB × IhG = MG,
hG =
dt
dt
(ext)
N
is the total external moment
where B is a body frame and MG =
i=1 ri/G × Fi
on the body.
Id
.
I
Qualitative analysis of a rapidly spinning rigid body is based on the approximation
Bd
dt
hG ≈ 0,
I
which simplifies the rotational dynamics to the vector cross product
I B
ω × IhG ≈ MG.
.
The moment of inertia tensor of a rigid body about its center of mass is
IG =
N
mi (ri/G . ri/G)U − (ri/G ⊗ ri/G) ,
i=1
where U is the unity tensor, and the rigid body is made up of N particles. For a
continuous rigid body B, the moment of inertia tensor is
(rdm/G . rdm/G)U − (rdm/G ⊗ rdm/G) dm.
IG =
B
.
The moment of inertia matrix of a rigid body about its center of mass is
[IG]B =
N
mi ([ri/G]TB [ri/G]B )I − [ri/G]B [ri/G]TB .
i=1
where I is the 3 × 3 identity matrix. For a continuous rigid body B, the moment
of inertia matrix is
([rdm/G]TB [rdm/G]B )I − [rdm/G]B [rdm/G]TB dm.
[IG]B =
B
.
A three-dimensional body frame can always be chosen so that the moment of
inertia matrix is diagonal. In this case, the axes of the body frame are called the
principal axes, and the nonzero entries in the moment of inertia matrix are called
the principal moments of inertia.
MULTIPARTICLE AND RIGID-BODY DYNAMICS IN THREE DIMENSIONS
.
525
The angular momentum of a rigid body about its center of mass is
I
hG = IG . IωB
or, in matrix form,
[IhG]B = [IG]B [IωB ]B .
.
The rotational equations of motion of a rigid body are written in terms of the
angular velocity of the body in absolute space:
IG .
.
Bd I B
dt
ω
+ IωB × IG . IωB = MG.
Euler’s equations are the three scalar rotational equations of motion of a rigid
body expressed in the principal axes of a body frame,
I1ω̇1 + (I3 − I2)ω2ω3 = M1
I2ω̇2 + (I1 − I3)ω1ω3 = M2
I3ω̇3 + (I2 − I1)ω1ω2 = M3,
where I1, I2, and I3 are the principal moments of inertia.
.
The parallel axis theorem in three dimensions provides a formula for computing
the moment of inertia tensor about an arbitrary point Q fixed to a rigid body:
IQ = IG + mG (rQ/G . rQ/G)U − (rQ/G ⊗ rQ/G)
or, in matrix form,
[IQ]B = [IG]B + mG rQ/G2I − [rQ/G]B [rQ/G]TB .
.
The moment of inertia matrix expressed as components in a body-fixed frame B1
can be rotated so that it is expressed as components in a different frame B2 by
means of
T
[IG]B2 = B2 C B1[IG]B1 B2 C B1 ,
where B2 C B1 is the direction-cosine matrix from B1 to B2.
.
The total kinetic energy of a three-dimensional multiparticle system about point
O separates into the kinetic energy of the center of mass G about O and the kinetic
energy about G:
TO = TG/O + TG,
where
1
TG/O = mGIvG/O 2
2
and
TG =
1I B . . I B 1I . I B
ω I G ω = hG ω .
2
2
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CHAPTER ELEVEN
.
The total energy of a three-dimensional rigid body is
EO =
TG/O + TG + UO
and the three-dimensional work-energy formula is
(nc)
EO (t2) = EO (t1) + WG/O
+ WG(nc).
11.10 Notes and Further Reading
Most introductory texts rely solely on a matrix approach when treating threedimensional rigid bodies and thus use only scalar equations of motion in the body
frame. We have chosen to use a tensor-based approach to be consistent with our
earlier vector treatment and to be able to formulate the equations of motion in frameindependent terms. Two other books that use a similar notation are Rao (2006) and
Tenenbaum (2004). A third notational approach, that of dyadics, is used throughout
Kane’s books (Kane 1978; Kane and Levinson 1985; Kane et al. 1983). Goldstein
(1980) also has a good discussion of tensors and dyadics. Readers interested in exploring more mathematically advanced approaches employing differential geometry
and group theory can consult O’Reilly (2008), Arnold (1989), or Marsden and Ratiu
(1999). (These texts also primarily use the energy-based methods that are introduced
in Chapter 13.)
As you might have surmised, we have only touched the surface here on the variety
of examples demonstrating rigid-body motion. For a slightly more advanced treatment
of torque-free motion (including the Poinsot construction for large-angle paths in
space) readers can consult Goldstein (1980), Greenwood (1988), or Hand and Finch
(1998). Goldstein (1980) and Hand and Finch (1998) also have excellent discussions
of the spinning Lagrange top. An excellent collection of aerospace examples (with
some applications to automatic control) can be found in Bryson (1994). Our ship
stabilization example is similar to one in that book. Three other references with good
discussions of space-vehicle dynamics are Kaplan (1976), Kane et al. (1983), and Wie
(1998). A good discussion of the rigid-body dynamics of an airplane can be found
in Stengel (2004). We did not discuss the topic of semi-rigid bodies, that is, bodies
with moving parts that change the moment of inertia. This greatly complicates the
analysis, but is important for some systems, such as a spacecraft with moving masses
for attitude control. One reference that touches on the topic is Kaplan (1976).
There is some confusion in the literature regarding the use of the terms precession
and nutation. Most texts follow the traditional approach of using precession to describe the circular motion of the body symmetry axis about an inertial direction (e.g.,
Fetter and Walecka 1980; Goldstein 1980; Greenwood 1988; Hand and Finch 1998).
To highlight the differences between torque-free and torqued bodies, we have chosen
to use the expression free-body precession to refer to this motion of a torque-free rigid
body. The nutation angle refers to the offset of the symmetry axis from an inertial direction and nutation refers to the oscillatory motion of this angle for spinning bodies
subject to torque. There are a number of books on spacecraft dynamics, however, that
use nutation to refer to the motion of the angular velocity about the angular momentum in torque-free motion (e.g., Wertz 1978; Bryson 1994). Thus nutation dampers,
such as the one described in Problem 11.19, damp out this motion of the angular
MULTIPARTICLE AND RIGID-BODY DYNAMICS IN THREE DIMENSIONS
527
velocity. We find this nomenclature confusing and have tried to consistently call this
motion precession; a nutation damper thus brings the nutation angle to zero.
11.11 Problems
11.1 How many degrees of freedom does a rigid airplane have while flying in the
air? While rolling on a runway?
11.2 Find the equations of motion for a gyropendulum where the spinning rotor
is a disk. Assume the rotor has fixed angular momentum h about its axis and
has a transverse moment of inertia J . That is, treat the spinning mass not as a
particle but as a rigid body.
11.3 Show that the three-dimensional parallel axis theorem in Eq. (11.56) reduces
to the planar theorem for rigid-body rotation in two dimensions (Eq. (9.34)).
11.4 Verify Eq. (11.77) for the kinetic energy about an inertially fixed point O (TO )
in terms of the kinetic energy, TQ, about an arbitrary point of the body Q and
the kinetic energy, TQ/O , due to translation of Q.
11.5 Find the angular velocity of the rigid two-wheel system shown in Figure 11.27,
assuming that both wheels roll without slipping at rate
(Assume the wheels are fixed to the axle.)
R2
R1
and R2 > R1.
Ω
L
Figure 11.27 Problem 11.5.
11.6 Recall the right circular cone of height h, radius R, and mass m of Example 11.1. Let O be the tip of the cone and assume it has uniform density.
a. Find the moment of inertia matrix about the center of mass G.
b. Find the moment of inertia matrix about O.
11.7 Example 11.6 determined the moment of inertia of a solid cube. Now you will
find the moment of inertia of a hollow cube, where the wall thickness is much
smaller than the cube height.
a. Consider a thin square plate of height h and mass mp . Show that the
moment of inertia of such a plate, relative to its center of mass G and
expressed in a body frame B attached to the plate, is
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CHAPTER ELEVEN
⎡
mp h 2
⎢ 12
⎢
⎢
[IG]B = ⎢
⎢ 0
⎢
⎣
0
⎤
0
0
mp h 2
12
0
⎥
⎥
⎥
0 ⎥
⎥.
⎥
mp h 2 ⎦
6
B
b. A hollow cube can be modeled as a structure built of six thin plates.
For one of the plates, find the moment of inertia relative to the center
of the cube O.
c. Sum the contributions of all the plates to find the moment of inertia of
the cube about its center of mass. Be careful to consider the rotation
of the plates with respect to one another.
11.8 Use a qualitative analysis to predict the motion of a frisbee that is inclined
at angle θ to the oncoming wind, as shown from the side in Figure 11.28.
How does this motion differ from what would happen if the frisbee were not
spinning?
I
hG
Wind
θ
Figure 11.28 Problem 11.8.
11.9 A three-dimensional top with a pointed tip fixed at O on a frictionless surface
spins rapidly about its long axis, as shown in Figure 11.29. Angle ψ is
determined by the projection of the center of mass G onto the horizontal plane.
The top is inclined at angle θ and spins about its long axis at rate φ̇ 0. Use
a qualitative description of the rotational dynamics to explain the gyroscopic
motion of the top. [HINT: Consider the rotational dynamics about O (i.e.,
Id
I
I
I
I
I
dt ( hO ) = MO ), and use the approximation hO = hG/O + hG ≈ hG .]
I
ϕ0
a3
G
g
a2
l
O
e3, a3
G
O
θ
a1
e1
ψ
a1
e2
Figure 11.29 Problem 11.9.
11.10 Use matlab to integrate the torque-free rigid-body rotational equations of
motion in Example 11.10 with I3 > I2 > I1 (i.e., Euler’s equations with M1 =
M2 = M3 = 0). For each of the following three sets of initial conditions, briefly
describe in words the resulting motion of the rigid body:
MULTIPARTICLE AND RIGID-BODY DYNAMICS IN THREE DIMENSIONS
529
a. ω1 > 1 and ω2 ≈ ω3 1,
b. ω2 > 1 and ω3 ≈ ω1 1, and
c. ω3 > 1 and ω1 ≈ ω2 1.
11.11 A rotary pendulum, shown in Figure 11.30, is composed of a uniform disk of
radius R and a thin uniform rod of length l and mass mr , pinned by its end to
the edge of the disk. The rod swings in the plane tangent to the edge of the
disk. Assuming the disk has a moment of inertia J about its center, find the
equations of motion of this system.
R
ψ
l
θ
Figure 11.30 Problem 11.11.
11.12 Find the equations of motion of a passenger on the amusement park ride shown
in Figure 11.31 by developing a physical model and analyzing it. (How many
degrees of freedom are in your model?)
Figure 11.31 Problem 11.12. Image courtesy of Shutterstock.
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CHAPTER ELEVEN
11.13 Consider again the demonstration gyro from Example 11.13. Instead of the
small mass having a fixed position, suppose it is connected to a spring of spring
constant k and unstretched length 0 that is attached to the post.
a. How many degrees of freedom does the system have?
b. Find the equations of motion for the entire system.
c. Suppose the gyro is spinning but otherwise balanced and an impulse is
applied to the small mass. What is the total angular momentum before
and after the impulse?
d. Let the mass of the gyro be 5 kg, the small mass be 2 kg, and set
l = 0.5 m and l = 0.25 m. The gyro is spinning at a rate of 1 rad/s,
and has the moment of inertia of a thin circular disk with a radius
of 10 cm. The spring has a constant of 0.1 N/m and a rest length of
zero. Numerically integrate the equations of motion after an impulse is
applied (you can choose the size of the impulse). Qualitatively describe
the resulting motion. Remember that the system is at rest before the
impulse is applied when choosing your initial conditions.
e. Suppose a damper is added in addition to the spring. What is the steadystate configuration of the system?
11.14 Figure 11.32 shows the basin of a typical washing machine filled with clothes.
The basin spins about its axis and rotates about the pivot point at its center.
When balanced, it stays level and simply spins about its axis. However, you
may have experienced imbalance during the spin cycle, when it sounds like the
basin will break through the machine. Figure 11.32b shows such a situation.
Ω
Ω
(a)
(b)
Figure 11.32 Problem 11.14. (a) Balanced load. (b) Unbalanced load.
a. How many degrees of freedom does the system have? Select a set of
scalar coordinates to describe the degrees of freedom.
b. Describe qualitatively the expected motion when the clothes are imbalanced.
11.15 Consider the upper stage of a long narrow rocket used to transfer a satellite
from a low-earth orbit to geosynchronous orbit (thus it is out of the atmosphere). Assume a small misalignment of the thrust vector FT from the center
line, as shown in Figure 11.33. For large high-thrust solid rocket motors, the
rockets are always spun about the long axis. Explain the reason for spinning
MULTIPARTICLE AND RIGID-BODY DYNAMICS IN THREE DIMENSIONS
531
I
ωB
FT
Figure 11.33 Problem 11.15.
the rocket stage. [HINT: Consider the rotational and translational motion of
the rocket when spinning and when not spinning.]
11.16 Suppose you take your new all-terrain vehicle (ATV) out for a spin, literally.
Assume you are traveling at a constant speed v, and you take a sharp circular
turn (either because it is fun or because you have to suddenly avoid an
obstacle). Unfortunately, you take the turn too fast and roll over (a common
problem).
a. Find the maximum speed that you can take a turn of radius of curvature
Rc before rolling over (the inside wheels6 leaving the ground) in terms
of L, the distance between the wheels, and d, the height above the
ground of the vehicle’s center of mass. Assume that the ATV travels
the turn without slipping; that is, the radius of curvature is constant.
b. Find the maximum speed for a typical ATV (with a wheelbase L of 3.5
ft and d of about 4 ft), assuming a sharp turn with radius of curvature
of 5 ft.
c. What changes might you make to the ATV design to prevent rollover?
B
θ
L
O
A
ϕ
Cord
Figure 11.34 Problem 11.17.
11.17 Consider a slender rod AB of mass m and length L pinned at its midpoint O
to a vertical shaft, as shown in Figure 11.34. The rod can rotate by θ about the
pin, and the shaft can rotate by φ about its long axis. The mass of the shaft
6 The inside wheels are the two wheels on the side of the car inside the turn, that is, toward the center of
the circular arc.
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CHAPTER ELEVEN
and collar may be neglected. The cord keeps the rod at an initial angle of 30◦
with the vertical while the shaft is initially rotating freely at angular velocity
φ̇ = 200 rad/s. The cord suddenly breaks.
a. Find an expression for the tension in the cord before it breaks.
b. Find the differential equations of motion for the rod (θ̈ and φ̈) after the
break.
11.18 Show that the most general parallel axis theorem for transporting the moment
of inertia tensor from an arbitrary point P of the body to a different arbitrary
point Q of the body is
(rQ/P . rQ/P )U + 2(rP/G . rQ/P )U
IQ = IP + mG
,
−rQ/P ⊗ rQ/P − rQ/P ⊗ rP/G − rP/G ⊗ rQ/P
where, as usual, G refers to the center of mass of the rigid body.
11.19 Consider a torque-free cylindrical satellite spinning about its major axis,
as shown in Figure 11.35. If the angular momentum, angular velocity, and
symmetry (spin) axis are all aligned, it will spin uniformly about the b2 axis,
ensuring that all instruments stay pointed. However, should it be disturbed (by
a micrometeorite impact, perhaps) so that the angular momentum is moved,
the satellite body frame will precess, as seen in Tutorial 11.1. Many satellites
are equipped with what is called a nutation damper to remove the small
nonspin-axis angular-velocity components. This device consists of a small
wheel free to spin about its symmetry axis at rotation rate but fixed to the
satellite about the other two axes, as shown in Figure 11.35. The wheel is
equipped with a viscous damper, so that it is subjected to a moment about its
spin axis of magnitude −D .
Suppose the satellite (including the wheel) has principal moments of inertia
(IT , IS , IT ), where IT < IS , and the wheel has a moment of inertia about its
b1
B
Ω
I
hG
G
b2
Figure 11.35 Problem 11.19.
b3
MULTIPARTICLE AND RIGID-BODY DYNAMICS IN THREE DIMENSIONS
533
spin axis of IW . Find the equations of motion for the three angular-velocity
components of the satellite, ω1, ω2, and ω3, and the spin rate of the wheel.
Using initial conditions (0) = 0, ω1(0) = 0.2 rad/s, ω2 = 2 rad/s, ω3(0) = 0,
IS /IT = 1.5, IW /IT = 0.06, and D = 0.5, show, numerically, that the system
does indeed damp the initial precession.
11.20 Consider the spinning, torque-free cylindrical satellite in Figure 11.36. Similar
to Problem 11.19, the cylinder is symmetric, with moment of inertia J about
b2 and moment of inertia I about b1 and b3. Note that J > I . The cylinder is
shown with an arbitrary angular velocity in the body frame. We would like
the cylinder to be spinning about the b2 axis, its symmetry axis (the axis
of maximum moment of inertia). As you know, if the cylinder has angular
velocity components in the other directions (b1 and b3), it will precess. As in
Problem 11.19, we have added a nutation damper.
The nutation damper here consists of a particle of mass m connected to
a spring of spring constant k and a viscous damper of damping constant b.
The particle is free to move only in the b2 direction. Let y represent the
displacement of the particle from the center of mass in the b2 direction.
The spring is unstretched when y = 0. The mass-spring-damper system is
displaced a distance l in the b3 direction from the center of mass of the cylinder.
Note that the mass properties of the cylinder do not include the nutation
damper.
For all parts you may assume that the center of mass of the cylinder is either
stationary in inertial space or moving at constant velocity.
a. Draw free-body diagrams for the cylinder and the particle. Include all
forces and moments (including constraint forces!).
b. Find the equations of motion for the angular velocity components of
the cylinder and the displacement of the particle. You may assume that
the cylinder is nominally spinning about b2 at ωs and that ω1, ω3, and
B
b3
ωB
I
G
b1
Figure 11.36 Problem 11.20.
b2
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CHAPTER ELEVEN
y, the displacement of the mass, can all be treated as small quantities
(thus higher powers and products among them can all be dropped).
Show that the resulting equations of motion are
ω̇1 − ω3 = −
ml 2
ml
ωn − ωs2 y − 2 ζ ωnẏ
I
I
ω̇3 + ω1 = 0
ÿ + 2ζ ωnẏ + ωn2 y + lωs ω3 = l ω̇1,
2
where = J −I
I ωs , ωn = k/m, and 2ζ ωn = b/m.
c. Using matlab, numerically integrate the equations of motion over
the interval [0, 40] s and plot ω1(t) and ω3(t) on a single plot and
y(t) on a second plot. Explain your results. (That is, does the design
work?)Use J = 1000 kg-m2, I = 200 kg-m2, m = 25 kg, l = 1 m,
ωn = k/mP = 1 rad/s, and ζ = 0.9. Assume that particle starts at
rest (y(0) = ẏ(0) = 0) and the cylinder has initial angular velocity
components ω1(0) = 0.1 rad/s, ω2(0) = ωs = 1 rad/s, and ω3(0) =
−0.2 rad/s.
d. What do you think would happen if the cylinder were long and thin
rather than short and fat (that is, J < I )?
11.21 Repeat Problem 11.19 with IT > IS . That is, suppose the satellite is a minor-
axis spinner with a nutation damper. Repeat the simulation with IT /IS = 1.5.
What do you notice? Does this result make sense?
PART FIVE
Advanced Topics
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CHAPTER TWELVE
Some Important Examples
This chapter applies the tools and methods developed in the first four parts of the
book to examine three important examples in a bit more depth. In the process, we
introduce some key concepts in dynamical systems that may be important if you
continue in more advanced studies. The first section explores simple harmonic motion
and vibrations, introducing you to the solution of multiple-degree-of-freedom linear
systems and mode shapes. The second section explores the ideas of equilibrium and
linearization to show how the behavior of a complex dynamical system can often be
predicted by a set of linear equations of motion. As an example, we return to the
airplane and present the linearized motion about a particular steady-state solution.
The final section returns to the topic of collisions introduced in Chapter 6 and applies
it to the collision of finite-sized particles.
12.1 An Introduction to Vibrations and Linear Systems
Chapter 2 introduced simple harmonic motion because of its importance and ubiquity.
Although the mechanical system solved in Tutorials 2.3 and 2.4 was a simple onedimensional model of a mass on a spring, by now you may have noticed how often
simple harmonic motion appears in dynamics. For example, the blades of a wind
turbine may vibrate due to wind gusts (as in Figure 12.1). When we talk about
vibrations, we simply mean the response of a dynamical system described by either
Eq. (2.20) or Eq. (2.24). For example, when we considered the solution of the
pendulum for small angles, we were able to convert its nonlinear equation of motion
to simple harmonic form. The importance of the solution to this differential equation
lies in its application to so many systems!
This section reviews the solution of simple harmonic motion and expands it to the
case of forced oscillations. We then show how this solution can be applied to any linear
dynamic system by generalizing the solution approach. We conclude by introducing
higher degree-of-freedom systems incorporating coupled oscillators.
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CHAPTER TWELVE
Figure 12.1 Vibrations of the blades of a wind turbine exhibit simple harmonic motion.
12.1.1 A Review of Simple Harmonic Motion
Here we briefly review the homogeneous solution from Chapter 2 for the undamped
and damped simple harmonic oscillator. By homogenous we mean the solution of the
equations of motion for certain initial conditions with no other forces acting. We are
looking for the solution of
z̈ + ω02z = 0
(12.1)
z̈ + 2ζ ω0ż + ω02z = 0
(12.2)
for the undamped case, and of
for the damped case (ω0 is the natural frequency and ζ is the damping ratio). Recall
that we found the solution by recognizing that the general solution must be the sum
of two complex exponentials,
z(t) = Az1(t) + Bz2(t),
where z1(t) = eλ1t and z2(t) = eλ2t for some λ1, λ2, and A and B are constants given
by the initial conditions. Substituting z(t) into Eq. (12.2) shows that λ1 and λ2 must
be solutions to the characteristic equation:
λ2 + 2ζ ω0λ + ω02 = 0,
(12.3)
SOME IMPORTANT EXAMPLES
539
which has the two roots
λ1,2 = −ζ ω0 ± ω0 ζ 2 − 1.
For the undamped case (ζ = 0), the roots are λ1,2 = ±ω0i and the response is
sinusoidal. Using eiω0t = cos ω0t + i sin ω0t and the initial conditions yields
z(t) = z(0) cos(ω0t) +
ż(0)
sin(ω0t)
ω0
ż(t) = −z(0)ω0 sin(ω0t) + ż(0) cos(ω0t).
The important point here is that the constant ω0 is the natural frequency of oscillation
of the solution. Figure 2.8 shows this sinusoidal response.
For the damped case (ζ > 0), we found three types of solutions: (a) an underdamped solution, where 0 < ζ < 1; (b) a critically damped solution, where ζ = 1; and
(c) an overdamped solution, where ζ > 1. The underdamped
solution is a decaying
ω0 1 − ζ 2 :
sinusoidal oscillation with damped frequency ωd =
z(t) = e−ζ ω0t c1 cos(ωd t) + c2 sin(ωd t) ,
where c1 and c2 are constants that depend upon the initial conditions:
c1 = z(0)
c2 =
ż(0) + z(0)ζ ω0
.
ω0 1 − ζ 2
The amplitude of the underdamped oscillations remains inside a decaying envelope
given by e−ζ ω0t . The underdamped solution is plotted in Figure 2.9. Note also that, if
ζ = 0, we recover the undamped solution.
The overdamped solution is characterized by an exponential decay with no
oscillation,
√
√
−ω0 ζ + ζ 2 −1 t
−ω0 ζ − ζ 2 −1 t
+ Be
,
z(t) = Ae
where A and B are found from the initial conditions to be
ż(0) + z(0)ω0 ζ − ζ 2 − 1
A=−
2ω0 ζ 2 − 1
ż(0) + z(0)ω0 ζ + ζ 2 − 1
B=
.
2ω0 ζ 2 − 1
The overdamped solution is also plotted in Figure 2.9.
Finally, the critically damped solution (ζ = 1) is the boundary between the overdamped and underdamped solutions. It is also characterized by an exponential decay,
though the general solution is z(t) = Ae−iζ ω0t + Bte−iζ ω0t :
z(t) = z(0)e−ζ ω0t + (ż(0) + z(0)ζ ω0)te−ζ ω0t .
It is also shown in Figure 2.9.
540
CHAPTER TWELVE
It is also interesting to look at the total energy EO of the simple harmonic oscillator.
Using the solution for the underdamped case and E0 = 21 mż2 + 21 kz2, we find
2
1 EO = m −ζ ω0e−ζ ω0t (c1 cos ωd t + c2 sin ωd t) + ωd e−ζ ω0t (−c1 sin ωd t + c2 cos ωd t)
2
1
+ ke−2ζ ω0t (c1 cos ωd t + c2 sin ωd t)2.
2
Using ω02 = k/m, this expression can be simplified to
1
1
EO = e−2ζ ω0t k(c12 + c22) + ζ 2ke−2ζ ω0t (c12 − c22) cos(2ωd t) + 2c1c2 sin(2ωd t)
2
2
1
+ mζ ωd e−2ζ ω0t (c12 − c22) sin 2ωd t − 2c1c2 cos 2ωd t .
2
As expected, the total energy is a constant of the motion for ζ = 0. For the damped
case, the damping coefficient gives the rate at which energy is removed from the
system, that is, the rate of nonconservative work done by the damper.
Remember, the solution of any dynamical system for which the equation of motion
can be reduced to either Eq. (12.1) or Eq. (12.2) will show this behavior. Of course,
any system that we can model by means of a mass on a spring will respond this way!
12.1.2 Forced Response and Resonance
This section examines forced vibrations. Consider a mass on a spring with rest length
x0. What form does the solution take when we allow an arbitrary force to act on the
mass, as shown in Figure 12.2? The resulting equation of motion is
ẍ + 2ζ ω0ẋ + ω02x =
−FP (t)
,
mP
(12.4)
where x is the displacement of the mass from the rest position.
The general solution of Eq. (12.4) is beyond the scope of this text. However, there
are three specific types of forces considered here that have important, and ubiquitous,
solutions: (a) an impulsive force, (b) a constant force, and (c) a sinusoidal driving
I
ey
x
FP (t)
–kxex
mP
ex
Figure 12.2 Mass on a spring with an arbitrary applied force in the −ex direction.
SOME IMPORTANT EXAMPLES
541
force. Example 4.1 examined the impulsive force. We found that considering an
impulsive force on the mass is equivalent to finding the unforced, or homogenous,
solution with initial conditions given by Eqs. (4.3) and (4.4). We also looked at the
constant-force solution toward the end of Tutorial 2.3. Nevertheless, let us revisit that
case, as it introduces the process we will use for more complicated forces.
In general, when searching for the solution of a driven or inhomogeneous equation
of motion, such as Eq. (12.4), we write it as a sum of two parts, the homogenous
solution and the particular solution:
x(t) = xh (t) + xp(t).
The homogeneous solution is the sinusoidal response already found for FP (t) = 0.
This general solution must be correct, since it satisfies the equation of motion. At this
point the linearity of the differential equation becomes relevant. It is only because the
equation of motion is linear in x that we can write the general solution as the sum
of these two parts. Thus the response due to initial conditions is entirely contained
in xh (t). To find the forced response, we just have to guess the right xp(t) due to
the forcing function. For a constant force, that is something we have already seen—
xp(t) = xp(c) is a constant. Plugging into the equation of motion with FP = F0 gives
xp =
−F0
.
mP ω02
Thus the general solution of the underdamped case with a constant driving force
becomes
x(t) =
−F0
+ e−ζ ω0t c1 cos(ωd t) + c2 sin(ωd t) .
2
mP ω0
(12.5)
Example 12.1 Bungee Jump
This example looks at the oscillatory solution of a mass on a spring-damper under
the weight of gravity. Assume